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Page No 74:
Question 1:
Which of the following expressions are polynomials? In case of a polynomial, write its degree.
(i) x5-2x3+x+3
(ii) y3+3y
(iii) t2-25t+5
(iv) x100-1
(v) 12x2-2x+2
(vi) x-2+2x-1+3
(vii) 1
(viii) -35
(ix) x22-2x2
(x) 23x2-8
(xi) 12 x2
(xii) 15x12+1
(xiii) 35x2-73x+9
(xiv)
x4-x32+x-3
(xv) 2x3+3x2+x-1
Answer:
(i) x5-2x3+x+3 is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 5, so, it is a polynomial of degree 5.
(ii) y3+ 3y is an expression having only non-negative integral powers of y. So, it is a polynomial. Also, the highest power of y is 3, so, it is a polynomial of degree 3.
(iii) t2-25t+5 is an expression having only non-negative integral powers of t. So, it is a polynomial. Also, the highest power of t is 2, so, it is a polynomial of degree 2.
(iv) x100-1 is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 100, so, it is a polynomial of degree 100.
(v) 12x2-2 x+2 is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.
(vi) x- 2+2x-1+3 is an expression having negative integral powers of x. So, it is not a polynomial.
(vii) Clearly, 1 is a constant polynomial of degree 0.
(viii) Clearly, -35 is a constant polynomial of degree 0.
(ix) x22-2x2=x22-2x-2
This is an expression having negative integral power of x i.e. −2. So, it is not a polynomial.
(x) 23x2-8 is an expression having only non-negative integral power of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.
(xi) 12x2= 12x-2 is an expression having negative integral power of x. So, it is not a polynomial.
(xii) 15x12+1
In this expression, the power of x is 12 which is a fraction. Since it is an expression having fractional power
of x, so, it is not a polynomial.
(xiii) 35x2-73x+9 is an expression having only non-negative integral powers of x. So, it is a polynomial. Also, the highest power of x is 2, so, it is a polynomial of degree 2.
(xiv) x4-x32+x-3
In this expression, one of the powers of x is 3 2 which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.
(xv) 2x3+3x2+x-1=2x3+3x2+
x12-1
In this expression, one of the powers of x is 12 which is a fraction. Since it is an expression having fractional power of x, so, it is not a polynomial.
Page No 75:
Question 2:
Identify constant, linear, quadratic, cubic and quartic polynomials from the following.
(i) –7 + x
(ii) 6y
(iii) –z3
(iv) 1 – y – y3
(v) x – x3 + x4
(vi) 1 + x + x2
(vii) – 6x2
(viii) – 13
(ix) – p
Answer:
(i) –7 + x is a polynomial with degree 1. So, it is a linear polynomial.
(ii) 6y is a polynomial with degree 1. So, it is a linear polynomial.
(iii) –z3 is a polynomial with degree 3. So, it is a cubic polynomial.
(iv) 1 – y – y3 is a polynomial with degree 3. So, it is a cubic polynomial.
(v) x – x3 + x4 is a polynomial with degree 4. So, it is a quartic polynomial.
(vi) 1 + x + x2 is a polynomial with degree 2. So, it is a quadratic polynomial.
(vii) – 6x2 is a polynomial with degree 2. So, it is a quadratic polynomial.
(viii) –13 is a polynomial with degree 0. So, it is a constant polynomial.
(ix) – p is a polynomial with degree 1. So, it is a linear polynomial.
Page No 75:
Question 3:
Write
(i) the coefficient of x3 in x+3x2-5x3+x4.
(ii) the coefficient of x in 3- 22x+6x2.
(iii) the coefficient of x2 in 2x – 3 + x3.
(iv) the coefficient of x in 38x2-27x+16.
(v) the constant term in π2x2+7x-25π.
Answer:
(i) The coefficient of x3 in x+3x2-5x3+ x4 is −5.
(ii) The coefficient of x in 3-22x+6x2 is -22.
(iii) 2x – 3 + x3 = – 3 + 2x + 0x2 + x3
The coefficient of x2 in 2x – 3 + x3 is 0.
(iv) The coefficient of x in 38x2-27x+16 is -27.
(v) The constant term in π2x2+7x-25π is -25π.
Page No 75:
Question 4:
Determine the degree of each of the following polynomials.
(i) 4x-5x2+6x32x
(ii) y2(y – y3)
(iii) (3x – 2) (2x3 + 3x2)
(iv) -12x+3
(v) – 8
(vi) x–2(x4+ x2)
Answer:
(i) 4x-5x2 +6x32x=4x2x-5x22x+6x32x=2-52x +3x2
Here, the highest power of x is 2. So, the
degree of the polynomial is 2.
(ii) y2(y – y3) = y3 – y5
Here, the highest power of y is 5. So, the degree of the polynomial is 5.
(iii) (3x – 2)(2x3 + 3x2) = 6x4 + 9x3 – 4x3 – 6x2 = 6x4 + 5x3 – 6x2
Here, the highest power of x is 4. So, the degree of the polynomial is 4.
(iv) -12x+3
Here, the highest power of x is 1. So, the degree of the polynomial is
1.
(v) – 8
–8 is a constant polynomial. So, the degree of the polynomial is 0.
(vi) x–2(x4 + x2) = x2 + x0
= x2 + 1
Here, the highest power of x is 2. So, the degree of the polynomial is 2.
Page No 75:
Question 5:
(i) Give an example of a monomial of degree 5.
(ii) Give an example of a binomial of degree 8.
(iii) Give an example
of a trinomial of degree 4.
(iv) Give an example of a monomial of degree 0.
Answer:
(i) A polynomial having one term is called a monomial. Since the degree of required monomial is 5, so the highest power of x in the monomial should be 5.
An example of a monomial of degree 5 is 2x5.
(ii) A polynomial having two terms is called a binomial. Since the degree of required binomial is 8, so the highest power of x in the binomial should be 8.
An example of a binomial of degree 8 is 2x8 − 3x.
(iii) A polynomial having three terms is called a trinomial. Since the degree of required trinomial is 4, so the highest power of x in the trinomial should be 4.
An example of a trinomial of degree 4 is 2x4 − 3x + 5.
(iv) A polynomial having one term is called a monomial. Since the degree of required monomial is 0, so the highest power of x in the monomial should be 0.
An
example of a monomial of degree 0 is 5.
Page No 75:
Question 6:
Rewrite each of the following polynomials in standard form.
(i) x-2x2+8+5x3
(ii) 23+4y2-3y+2y3
(iii) 6x3+2x-x5-3x2
(iv) 2+t-3t3+t4-t2
Answer:
A polynomial written either in ascending or descending powers of a variable is called the standard form of a polynomial.
(i) 8+x-2x2+5x3 is a polynomial in standard form as the powers of x are in ascending order.
(ii) 23-3y+4y2+2y3 is a polynomial in standard form as the powers of y are in ascending order.
(iii) 2x-3x2+6x3 -x5 is a polynomial in standard form as the powers of x are in ascending order.
(iv) 2+t-t2-3t3+t4 is a polynomial in standard form as the powers of t are in ascending order.
Page No 78:
Question 1:
If p(x) = 5 − 4x + 2x2, find (i) p(0), (ii) p(3), (iii) p(−2)
Answer:
i px=5-4x+2x2
⇒p0=5-4×0+2×02
=5-0+0=5
ii px=5-4x+2x2⇒p3=5-4×3+2×32
=5-12+18=11
iii px=5-4x+2x2⇒p-2=5-4×-2+2×-22
=5+8+8=21
Page No 78:
Question 2:
If p(y) = 4 + 3y − y2 + 5y3, find (i) p(0), (ii) p(2), (iii) p(−1).
Answer:
i py=4+3y-y2+5y3⇒p0=4+3×0-02 +5×03
=4+0-0+0=4
ii py= 4+3y-y2+5y3⇒p2=4+3×2-22+5×23
=4+6-4+40=46
iii py=4+3y-y2+5y3 ⇒p-1=4+3×-1--12+5×-13
=4-3-1-5=-5
Page No 78:
Question 3:
If f(t) = 4t2 − 3t + 6, find (i) f(0), (ii) f(4), (iii) f(−5).
Answer:
i ft=4t2-3t+6⇒f0=4×02-3×0+6
=0-0+6=6
ii ft=4t2-3t+6⇒f4=4×42-3×4+6
=64-12+6=58
iii ft=4t2-3t+6⇒f-5=4×-52-3 ×-5+6
=100+15+6=121
Page No 78:
Question 4:
If px =x3-3x2+2x, find p(0), p(1), p(2). What do you conclude?
Answer:
px=x3-3x2+2x .....(1)
Putting x = 0 in (1), we get
p0=03-3×02+2×0=0
Thus, x = 0 is a zero of p(x).
Putting x = 1 in (1), we get
p1=13-3×12+2×1=1-3+2=0
Thus, x = 1 is a zero of p(x).
Putting x = 2 in (1), we get
p2=23-3×22+2×2=8-3×4+4=8-12+4=0
Thus, x = 2 is a zero of p(x).
Page No 78:
Question 5:
If p(x) = x3 + x2 – 9x – 9, find p(0), p(3), p(–3) and p(–1). What do you conclude about the zero of p(x)? Is 0 a zero of p(x)?
Answer:
p(x) = x3 + x2 – 9x – 9 .....(1)
Putting x = 0 in (1), we get
p(0) = 03 + 02 – 9 × 0 – 9 = 0 + 0 – 0 – 9 = –9 ≠ 0
Thus, x = 0 is not a zero of p(x).
Putting x = 3 in (1), we get
p(3) = 33 + 32 – 9 × 3 – 9 = 27 + 9 – 27 – 9 = 0
Thus, x = 3 is a zero of p(x).
Putting x = –3 in (1), we get
p(–3) = (–3)3 + (–3)2 – 9 × (–3) – 9 = –27 + 9 + 27 – 9 = 0
Thus, x = –3 is a zero of p(x).
Putting x = –1 in (1), we get
p(–1) = (–1)3 + (–1)2 – 9 × (–1) – 9 = –1 + 1 + 9 – 9 = 0
Thus, x = –1 is a zero of p(x).
Page No 78:
Question 6:
Verify that:
(i) 4 is a zero of the polynomial p(x) = x − 4.
(ii) −3 is a zero of the polynomial q(x) = x + 3.
(iii) 25is a zero of the polynomial, f(x) = 2 − 5x.
(iv) -12is a
zero of the polynomial g(y) = 2y + 1.
Answer:
i px=x-4⇒p 4=4-4
= 0
Hence, 4 is the zero of the given polynomial.
ii px=-3+3⇒p 3=0
Hence, 3 is the zero of the given polynomial.
iii px=2-5x⇒p25=2-5×25
=2-2=0
Hence, 25 is the zero of the given polynomial.
iv py =2y+1⇒p-12=2×-12+1 =-1+1 =0
Hence, -12 is the zero of the given polynomial.
Page No 79:
Question 7:
Verify that
(i) 1 and 2 are the zeros of the polynomial p(x) = x2 − 3x + 2.
(ii) 2 and −3 are the zeros of the polynomial q(x) = x2 + x − 6.
(iii) 0 and 3 are the zeros of the polynomial r(x) = x2 − 3x.
Answer:
i px=x2-3x+2=x-1x-2⇒p1=1-1×1-2
=0×-1=0
Also,
p2=2-12-2
=-1×0=0
Hence, 1 and 2 are the zeroes of the given polynomial.
ii px=x2+x-6⇒p2=22+2-6
=4-4=0
Also,
p-3=-32+-3-6 =9-9 =0
Hence, 2 and -3 are the zeroes of the given polynomial.
iii px=x2-3x⇒p0=02-3×0
Also,
p3=32-3×3 =9-9 =0
Hence, 0 and 3 are the zeroes of the given polynomial.
Page No 79:
Question 8:
Find the zero of the polynomial:
(i) p(x) = x − 5
(ii) q(x) = x + 4
(iii) p(t) = 2t − 3
(iv) f(x) = 3x + 1
(v) g(x) = 5 − 4x
(vi) h(x) = 6x − 1
(vii) p(x) = ax + b, a ≠ 0
(viii) q(x) = 4x
(ix) p(x) = ax, a ≠ 0
Answer:
i px=0⇒x-5=0 ⇒x=5Hence, 5 is the zero of the polynomial px.ii qx=0⇒x+4=0 ⇒x=-4 Hence, -4 is the zero of the polynomial qx.iii pt=0⇒2 t-3=0 ⇒t=32Hence, 3 2 is the zero of the polynomial pt.
iv fx=0⇒3x+1=0 ⇒x=-13Hence, -13 is the zero of the polynomial fx.v gx=0⇒5-4x=0 ⇒x=54Hence, 54 is the zero of the polynomial gx.vi hx=0⇒6x-1=0 ⇒x=16Hence, 16 is the zero of the polynomial hx.
vii px=0⇒ax+b=0 ⇒x=-baHence, -ba is the zero of the polynomial px.viii qx=0⇒4x=0 ⇒x=0Hence, 0 is the zero of the polynomial q x.ix px=0⇒ax=0 ⇒x=0Hence, 0 is the zero of the polynomial px.
Page No 79:
Question 9:
If 2 and 0 are the zeros of the polynomial fx=2x3-5x2+ax+b then find the values of a and b.
Hint f(2) = 0 and f(0) = 0.
Answer:
It is given that 2 and 0 are the zeroes of the polynomial fx=2x3-5x2+ax+b.
∴ f(2) = 0
⇒2×23-5×22+a ×2+b=0⇒16-20+2a+b=0⇒-4+2a+b=0⇒2a+b=4 .....1
Also,
f(0) = 0
⇒2×03-5×02+a×0+b=
0⇒0-0+0+b=0⇒b=0
Putting b = 0 in (1), we get
2a+0=4⇒2a=4⇒a=2
Thus, the values of a and b are 2 and 0, respectively.
Page No 84:
Question 1:
By actual division, find the quotient and the remainder when (x4 + 1) is divided by (x – 1).
Verify that remainder = f(1).
Answer:
Let f(x) = x4 + 1 and g(x) = x – 1.
Quotient = x3 + x2 + x + 1
Remainder = 2
Verification:
Putting x = 1 in f(x), we get
f(1) = 14 + 1 = 1 + 1 = 2 = Remainder, when f(x) = x4 + 1 is divided by g(x) = x – 1
Page No 84:
Question 2:
Verify the division algorithm for the polynomials px=2x4-6x3+2x 2-x+2 and gx=x+2.
Answer:
px=2x4-6x3+2x2-x +2 and gx=x+2
Quotient = 2x3-10 x2+22x-45
Remainder = 92
Verification:
Divisor × Quotient +
Remainder
=x+2×2x3-10x2+22x-45+ 92=x2x3-10x2+22x-45+22x3-10x2+22x-45+92 =2x4-10x3+22x2-45x+4x3-20x2+44x-90+92=2x4-6 x3+2x2-x+2
= Dividend
Hence verified.
Page No 84:
Question 3:
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=x3-6 x2+9x+3, gx=x-1.
Answer:
px=x3-6x2+9x+3
gx=x-1
By remainder theorem, when p(x) is divided by (x − 1), then the remainder = p(1).
Putting x = 1 in p(x), we get
p1=13-6× 12+9×1+3=1-6+9+3=7
∴ Remainder = 7
Thus, the remainder when p(x) is divided by g(x) is 7.
Page No 84:
Question 4:
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=2x3-7x2+9x-13, gx=x-3.
Answer:
p x=2x3-7x2+9x-13
gx=x-3
By remainder theorem, when p(x) is divided by (x − 3), then the remainder = p(3).
Putting x = 3 in p(x), we get
p3=2×33-7×32+9×3-13=54-63+27-13=5
∴ Remainder = 5
Thus, the remainder when p(x) is divided by g(x) is 5.
Page No 84:
Question 5:
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=3x4-6x2- 8x-2, gx=x-2.
Answer:
px=3x4-6x2-8x-2
gx=x-2
By remainder theorem, when p(x) is divided by (x − 2), then the remainder = p(2).
Putting x = 2 in p(x), we get
p2=3×24 -6×22-8×2-2=48-24-16-2=6
∴ Remainder = 6
Thus, the remainder when p(x) is divided by g(x) is 6.
Page No 84:
Question 6:
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=2x3-9x2+x+15, gx=2x-3.
Answer:
px=2x3-9x2+x+15
gx=2x-3=2x-32
By remainder theorem, when p(x) is divided by (2x − 3), then the remainder = p32.
Putting x=32 in p(x), we get
p 32=2×323-9×322+32+15=274-814+32+15 =27-81+6+604=124=3
∴ Remainder = 3
Thus, the remainder when p(x) is divided by g(x) is 3.
Page No 84:
Question 7:
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=x3-2x2-8x-1, gx=x+1.
Answer:
px=x3-2x2-8x-1
gx=x+1
By remainder theorem, when p(x) is divided by (x + 1), then the remainder = p(−1).
Putting x = −1 in p(x), we get
p-1=-13-2×-12-8×-1 -1=-1-2+8-1=4
∴ Remainder = 4
Thus, the remainder when p(x) is divided by g(x) is 4.
Page No 84:
Question 8:
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=2x3+x2-15x-12, gx=x+2.
Answer:
px=2 x3+x2-15x-12
gx=x+2
By remainder theorem, when p(x) is divided by (x + 2), then the remainder = p(−2).
Putting x = −2 in p(x), we get
p-2=2×-23+-22-15×-2-12=-16+4+30- 12=6
∴ Remainder = 6
Thus, the remainder when p(x) is divided by g(x) is 6.
Page No 84:
Question 9:
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=6x3 +13x2+3, gx=3x+2.
Answer:
px=6x3+13x2+3
gx=3x+2=3x+23=3x--23
By remainder theorem, when p(x) is divided by (3x + 2), then the remainder = p-23.
Putting x=-23 in p(x), we get
p-23 =6×-233+13×-232+3=-169+529+3 = -16+52+279=639=7
∴ Remainder = 7
Thus, the remainder when p(x) is divided by g(x) is 7.
Page No 84:
Question 10:
Using the remainder theorem, find the remainder, when
p(x) is divided by g(x), where
px=x3-6x2+2x-4, gx=1-32x.
Answer:
p x=x3-6x2+2x-4
gx=1-32x=-32x-23
By remainder theorem, when p(x) is divided by 1-32x, then the remainder = p23.
Putting x=23 in p(x), we get
p23=233-6×232+2×23-4=8 27-83+43-4 =8-72+36-10827=-13627
∴ Remainder = -13627
Thus, the remainder when p(x) is divided by g(x) is -13627.
Page No 84:
Question 11:
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px =2x3+3x2-11x-3, gx=x+12.
Answer:
px=2 x3+3x2-11x-3
gx=x+12=x--12
By remainder theorem, when p(x) is divided by x+12, then the remainder = p-12.
Putting x=- 12 in p(x), we get
p-12=2×-123+3×-122- 11×-12-3=-14+34+112-3 =-1+3+22-124=12 4=3
∴ Remainder = 3
Thus, the remainder when p(x) is divided by g(x) is 3.
Page No 84:
Question 12:
Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
px=x3 -ax2+6x-a, gx=x-a.
Answer:
px=x3-ax2+6x- a
gx=x-a
By remainder theorem, when p(x) is divided by (x − a), then the remainder = p(a).
Putting x = a in p(x), we get
pa =a3-a×a2+6×a-a=a3-a3+6a-a=5a
∴ Remainder = 5a
Thus, the remainder when p(x) is divided by g(x) is 5a.
Page No 84:
Question 13:
The polynomials 2x3+x2-ax+2 and 2x3-3x2-3x+ a when divided by (x – 2) leave the same remainder. Find the value of a.
Answer:
Let fx=2x3+x2-ax+2 and g x=2x3-3x2-3x+a.
By remainder theorem, when f(x) is divided by (x – 2), then the remainder = f(2).
Putting x = 2 in f(x), we get
f2 =2×23+22-a×2+2=16+4-2a+2=-2a+22
By remainder theorem, when g(x) is divided by (x – 2), then the remainder = g(2).
Putting x = 2 in g(x), we get
g2=2×23-3×22-3×2+a=16-12-6+a=-2+a
It is given that,
f2=g2⇒-2a+22=-2+a⇒-3a=-24⇒a=8
Thus, the value of a is 8.
Page No 84:
Question 14:
The polynomial p(x) = x4 − 2x3 + 3x2 − ax + b when divided by (x − 1) and (x + 1) leaves the remainders 5 and 19 respectively. Find the values of a and b. Hence, find the remainder when p(x) is divided by (x − 2).
Answer:
Let:px=x4-2x3+3x2-ax+b
Now,When p x is divided by x-1, the remainder is p1.When px is divided by x+1, the remainder is p-1.
Thus, we have:
p1=14-2 ×13+3×12-a×1+b =1-2+3-a+b =2-a+b
And,
p-1=-14-2×-13+3 ×-12-a×-1+b =1+2+3+a+b
=6+a+b
Now,
2-a+b=5 ...1 6+a+b=19 ...2
Adding (1) and (2), we get:8+2b=24
⇒2b=16⇒b=8
By putting the value of b, we get the value of a, i.e., 5.
∴ a = 5 and b = 8
Now,
fx = x4-2x3 + 3x2 - 5x + 8
Also,
When px is divided by x-2, the remainder is p2.
Thus, we have:
p2=24-2×2 3+3×22-5×2+8 a=5 and b=8 = 16-16+12-10+8 =10
Page No 85:
Question 15:
If px=x3-5 x2+4x-3 and gx=x-2, show that p(x) is not a multiple of g(x).
Answer:
px=x3-5 x2+4x-3
gx=x-2
Putting x = 2 in p(x), we get
p2=23-5×22 +4×2-3=8-20+8-3=-7≠0
Therefore, by factor theorem, (x − 2) is not a factor of p(x).
Hence, p(x) is not a multiple of g(x).
Page No 85:
Question 16:
If px=2x3-11x2-4x+5 and gx=2x+1, show that g(x) is not a factor of p(x).
Answer:
px=2x3-11x2-4x+5
gx=2x+1=2x+12 =2x--12
Putting x=-12 in p(x), we get
p-12 =2×-123-11×-122-4×-12+5=-14-114 +2+5 =-124+7=-3+7=4≠0
Therefore, by factor theorem, (2x + 1) is not a factor of p(x).
Hence, g(x) is not a factor of p(x).
Page No 90:
Question 1:
Using factor theorem, show that:
(x − 2) is a factor of (x3 − 8)
Answer:
Let:
fx=x3-8
Now,
x-2=0⇒x= 2
By the factor theorem, (x3- 8) is a factor of the given polynomial if f(2) = 0.
Thus, we have:
f2=23-8 =0
Hence, (x- 2) is a factor of the given polynomial.
Page No 90:
Question 2:
Using factor theorem, show
that:
(x − 3) is a factor of (2x3 + 7x2 − 24x − 45)
Answer:
Let:
f x=2x3+7x2-24x-45
Now,
x-3=0⇒x=3
By the factor theorem, (x - 3) is a factor of the given polynomial if f(3) = 0.
Thus, we have:
f3=2×33-7×32-24×3-45 = 54+63-72-45 =0
Hence, (x - 3) is a factor of the given polynomial.
Page No 90:
Question 3:
Using factor theorem, show that:
(x − 1) is
a factor of (2x4 + 9x3 + 6x2 − 11x − 6)
Answer:
Let:
fx=2x4+9x3+6x2-11x-6
Here,
x-1=0⇒x=1
By the factor theorem, (x - 1) is a factor of the given polynomial if f(1) = 0.
Thus, we have:
f1=2× 14+9×13+6×12-11×1-6 =2+9+6-11-6 =0
Hence, (x - 1) is a factor of the given polynomial.
Page No 91:
Question 4:
Using factor theorem, show
that:
(x + 2) is a factor of (x4 − x2 − 12)
Answer:
Let:
fx=x4-x2-12
Here,
x+2=0⇒x=-2
By the factor theorem, (x + 2) is a factor of the given polynomial if f (-2) = 0.
Thus, we have:
f-2=-24--22-12 =16-4-12 =0
Hence, (x + 2) is a factor of the given polynomial.
Page No 91:
Question 5:
Using factor theorem,
show that g(x) is a factor of p(x), when
px=69+11x-x2+x3, gx=x+3
Answer:
px=69+11x-x2+x3
gx=x+3
Putting x = −3 in p(x), we get
p-3=69+11×-3--32+-33=69-33-9-27=0
Therefore, by factor theorem, (x + 3) is a factor of p(x).
Hence, g(x) is a factor of p(x).
Page No 91:
Question 6:
Using factor theorem, show that:
(x + 5) is a factor of (2x3 + 9x2 − 11x − 30)
Answer:
Let:
fx=2x3+9x2-11x-30
Here,
x+5=0⇒x=-5
By the factor theorem, (x + 5) is a factor of the given polynomial if f (-5) = 0.
Thus, we have:
f-5=2×-53+9×-5 2-11×-5-30 =-250+225+55-30 =0
Hence, (x + 5) is a factor of the given polynomial.
Page No 91:
Question 7:
Using factor theorem, show that:
(2x − 3) is a factor of
(2x4 + x3 − 8x2 − x + 6)
Answer:
Let:
fx=2x4+x3-8x2-x+6
Here,
2x-3=0⇒x=32
By the factor theorem, (2x - 3) is a factor of the given polynomial if f32=0.
Thus, we have:
f32=2× 324+323-8×322-32+6 =818+278-18-32+6 =0
Hence, (2x - 3) is a factor of the given polynomial.
Page No 91:
Question 8:
Using factor
theorem, show that g(x) is a factor of p(x), when
px=3x3+x2-20x+12 ,gx=3x-2
Answer:
px=3x3+x2-20x+12
gx= 3x-2=3x-23
Putting x=23 in p(x), we get
p23=3× 233+232-20×23+12=89+49-403+12 =8+ 4-120+1089=120-1209=0
Therefore, by factor theorem, (3x − 2) is a factor of p(x).
Hence, g(x) is a factor of p(x).
Page No 91:
Question 9:
Using factor theorem, show that:
x-2is a factor of
7x2-42x-6
Answer:
Let:
fx= 7x2-42x-6
Here,
x-2=0⇒x=2
By the factor theorem, x-2 is a factor of the given polynomial if f2=0
Thus, we have:
f2=7×22-42×2 -6 =14-8-6 =0
Hence, x-2 is a factor of the given polynomial.
Page No 91:
Question 10:
Using factor theorem, show that:
x + 2 is a
factor of 22x2+5x +2
Answer:
Let:
fx=22x2+5x+ 2
Here,
x+2=0⇒x=-2
By the factor theorem, x+2 will be a factor of the given polynomial if f-2 = 0.
Thus, we have:
f-2=22×-22+5×-2 +2 =42-52+2 =0
Hence, x+2 is a factor of the given polynomial.
Page No 91:
Question 11:
Show that (p – 1) is a factor of (p10 – 1) and also of (p11 – 1).
Answer:
Let f(p) = p10 – 1 and g(p) = p11 – 1.
Putting p = 1 in f(p), we get
f(1) = 110 − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p10 – 1).
Now, putting p = 1 in g(p), we get
g(1) = 111 − 1 = 1 − 1 = 0
Therefore, by factor theorem, (p – 1) is a factor of (p11 – 1).
Page No 91:
Question 12:
Find the value of k for which (x − 1) is a factor of (2x3 + 9x2 + x + k).
Answer:
Let:
fx=2x3+9x2+x+k
x-1 is a factor of fx=2x3+9x2+x+k.⇒f1=0⇒2×13+9×12+1 +k=0⇒12+k=0⇒k=-12
Hence, the required value of k is -12.
Page No 91:
Question 13:
Find the value of a for which (x − 4) is a factor of (2x3 − 3x2 − 18x + a).
Answer:
Let:
fx=2x3-3x2-18x+a
x-4 is a factor of fx=2x3-3x2-18x+a.⇒f4=0⇒2×4 3-3×42-18×4+a = 0⇒8+a=0⇒a=-8
Hence, the required value of a is -8.
Page No 91:
Question 14:
Find the value of a for which (x + 1) is a factor of (ax3+ x2 – 2x + 4a – 9).
Answer:
Let f(x) = ax3 + x2 – 2x + 4a – 9
It is given that (x + 1) is a factor of f(x).
Using factor theorem, we have
f(−1) = 0
⇒a×-13+-12-2×-1+4a-9=0 ⇒-a+1+2+4a-9=0⇒3a-6=0⇒3a=6⇒a=2
Thus, the value of a is 2.
Page No 91:
Question 15:
Find the value of a for which (x + 2a) is a factor of (x5– 4a2 x3 + 2x + 2a +3).
Answer:
Let f(x) = x5 – 4a2 x3 + 2x + 2a +3
It is given that (x + 2a) is a factor of f(x).
Using factor theorem, we have
f(−2a) =
0
⇒-2a5-4a2×-2a3+2×-2a +2a+3=0⇒-32a5-4a2×-8a3+2×-2a+2a+3=0 ⇒-32a5+32a5-4a+2a+3=0⇒-2a+3=0
⇒2a=3⇒a=32
Thus, the value of a is 32.
Page No 91:
Question 16:
Find the value of m for which (2x – 1) is a factor of (8x4+ 4x3 – 16x2 + 10x + m).
Answer:
Let f(x) = 8x4 + 4x3 – 16x2 + 10x + m
It is given that 2x-1=2x-12 is a factor of f(x).
Using factor theorem, we have
f12=0⇒8×124+4×123-16×
122+10×12+m=0⇒12+12-4+5+m=0
⇒2+m= 0⇒m=-2
Thus, the value of m is −2.
Page No 91:
Question 17:
Find the value of a for which the polynomial (x4 − x3 − 11x2 − x + a) is divisible by (x + 3).
Answer:
Let:
fx=x4-x3-11x2-x+a
Now,
x+3=0⇒x=-3
By the factor theorem, fx is exactly divisible by x+3 if f-3=0.
Thus, we have:
f-3 =-34--33-11×-32--3+a =81+27-99+3+a =12+a
Also,
f-3=0⇒12+a=0⇒a=-12
Hence, fx is exactly divisible by x +3 when a is -12.
Page No 91:
Question 18:
Without actual division, show that (x3 − 3x2 − 13x + 15) is exactly divisible by (x2 + 2x − 3).
Answer:
Let:
fx=x3-3x2-13x+15
And,
gx=x2+2x-3
=x2+x-3x-3=xx-1+3x-1=x-1x+ 3
Now, fx will be exactly divisible by gx if it is exactly divisible by x-1 as well as
x+3.
For this, we must have:
f1=0 and f-3=0
Thus, we have:
f1=1 3-3×12-13×1+15 =1-3-13+15 =0
And,
f-3=-33-3×-32-13×-3+ 15 =-27-27+39+15 =0
fx is exactly divisible by x-1 as well as x+3. So, fx is exactly divisible by x-1x+3.
Hence, fx is exactly divisible by x2+2x- 3.
Page No 91:
Question 19:
If (x3 + ax2 + bx + 6) has (x − 2) as a factor and leaves a remainder 3 when divided by (x − 3), find the values of a and b.
Answer:
Let:
fx=x3+a x2+bx+6
x-2 is a factor of fx=x3+ax2+bx+6.⇒f2=0⇒23+a×22+b×2+6=0⇒14+4a+2b=0⇒ 4a+2b=-14⇒2a+b=-7 ...1
Now,
x-3 =0⇒x=3
By
the factor theorem, we can say:
When fx will be divided by x-3, 3 will be its remainder.⇒f3=3
Now,
f3=33+a×32+b×3+6 =27+9a+3b+6 =33+9a+3b
Thus, we have:
f 3=3⇒33+9a+3b=3⇒9a+3b=-30⇒3a+b=-10 ...2
Subtracting (1) from (2), we get:
a = -3
By putting the value of a in (1), we get the value of
b, i.e., -1.
∴ a = -3 and b = -1
Page No 91:
Question 20:
Find the values of a and b so that the polynomial (x3 − 10x2 + ax + b) is exactly divisible by (x −1) as well as (x − 2).
Answer:
Let:
f x=x3-10x2+ax+b
Now,
x-1=0⇒x=1
By the factor theorem, we can say:
fx will be exactly divisible by x-1 if f1=0.
Thus, we have:
f1=13-10×12+a×1+b =1-10+a+b =-9+a+b
∴ f1=0⇒a+b=9 ...1
Also,
x-2=0⇒x =2
By the factor theorem, we can say:
fx will be exactly divisible by x-2 if f2=0.
Thus, we have:
f2=23-10×22+a×2+b =8-40+2a+b =-32+2a+b
∴ f2=0⇒2a+b=32 ... 2
Subtracting (1) from (2), we get:a=23
Putting the value of a, we get the value of b, i.e., -14.
∴ a = 23 and b = -14
Page No 91:
Question 21:
Find the values of a and b so that the polynomial (x4 + ax3 − 7x2 − 8x + b) is exactly divisible by (x + 2) as well as (x + 3).
Answer:
Let:
fx=x4+ax3-7x2-8x+b
Now,
x+2=0⇒x=-2
By the factor theorem, we can say:
fx will be exactly divisible by x+2 if f-2=0.
Thus, we
have:
f-2 =-24+a×-23-7×-22-8×-2+b =16-8a-28+16+b =4-8a+ b
∴ f-2=0⇒8a-b=4 ...1
Also,
x+3 =0⇒x=-3
By the factor theorem, we can say:
fx will be exactly divisible by x+3 if f-3=0.
Thus, we have:
f-3=-34+a×-33-7×-32-8× -3+b =81-27a-63+24+b
=42-27a+b
∴ f-3=0⇒27a-b=42 ...2
Subtracting 1 from 2, we get:⇒19a=38⇒a=2
Putting the value of a, we get the value of b, i.e., 12.
∴ a = 2 and b = 12
Page No 91:
Question 22:
If both (x – 2) and x-12 are factors of px2 + 5x + r, prove that p = r.
Answer:
Let f(x) = px2 + 5x + r
It is given that (x – 2) is a factor of f(x).
Using factor theorem, we have
f2=0⇒p×22+5×2+r=0 ⇒4p+r=-10 .....1
Also, x-12 is a factor of f(x).
Using factor theorem, we have
f12=0⇒p×122+5×12+r=0 ⇒p4+r=-52⇒p+4r=-10 .....2
From (1) and (2), we have
4p+r=p+4r⇒4p-p=4r-r⇒3p=3r⇒p=r
Page No 91:
Question 23:
Without actual division, prove that 2x4 – 5x3 + 2x2 – x + 2 is divisible by x2 – 3x + 2.
Answer:
Let f(x) = 2x4 – 5x3 + 2x2 – x + 2 and g(x) = x2 – 3x + 2.
x2-3x+2=x2-2x-x+2=xx-2-1x-2 =x-1x-2
Now, f(x) will be divisible by g(x) if f(x) is exactly divisible by both (x − 1) and (x − 2).
Putting x = 1 in f(x), we get
f(1) = 2 × 14 – 5 × 13 + 2 × 12 – 1 + 2 = 2 – 5 + 2 – 1 + 2 = 0
By factor theorem, (x − 1) is a factor of f(x). So, f(x) is exactly divisible by (x − 1).
Putting x = 2 in f(x), we get
f(2) = 2 × 24 – 5 × 23 + 2 × 22 – 2 + 2 = 32 – 40 + 8 – 2 + 2 = 0
By factor theorem, (x − 2) is a factor of f(x). So, f(x) is exactly divisible by (x − 2).
Thus, f(x) is exactly divisible by both (x − 1) and (x − 2).
Hence, f(x) = 2x4 – 5x3 + 2x2 – x + 2 is exactly divisible by (x − 1)(x − 2) = x2 – 3x + 2.
Page No 91:
Question 24:
What must be added to 2x4– 5x3+ 2x2 – x – 3 so that the result is exactly divisible by (x – 2)?
Answer:
Let k be added to 2x4 – 5x3 + 2x2 – x – 3 so that the result is exactly divisible by (x – 2). Here, k is a constant.
∴ f(x) = 2x4 – 5x3 + 2x2 – x – 3 + k is exactly divisible by (x – 2).
Using factor theorem, we have
f2=0⇒2×24-5×23+2×22-2-3+k= 0⇒32-40+8-5+k=0⇒-5+k=0⇒k=5
Thus, 5 must be added to 2x4 – 5x3 + 2x2 – x – 3 so that the result is exactly divisible by (x – 2).
Page No 91:
Question 25:
What must be subtracted from (x4+ 2x3– 2x2 + 4x+ 6) so that the result is exactly divisible by (x2 + 2x – 3)?
Answer:
Dividing (x4 + 2x3 – 2x2 + 4x + 6) by (x2 + 2x – 3) using long division method, we have
Here, the remainder obtained is (2x + 9).
Thus, the remainder (2x + 9) must be subtracted from (x4 + 2x3 – 2x2 + 4x + 6) so that the result is exactly divisible by (x2 + 2x – 3).
Page No 91:
Question 26:
Use factor theorem to prove that (x + a) is a factor of (xn+ an) for any odd positive integer.
Answer:
Let f(x) = xn + an
Putting x = −a in f(x), we get
f(−a) = (−a)n + an
If n is any odd positive integer, then
f(−a) = (−a)n + an = −an + an = 0
Therefore, by factor theorem, (x + a) is a factor of (xn + an) for any odd positive integer.
Page No 92:
Question 1:
Which of the following expressions is a polynomial in one variable?
(a) x+ 2x+3
(b) 3x+2x+5
(c) 2x2 - 3x + 6
(d) x10 +y5 + 8
Answer:
(c) 2x2 - 3x + 6
Clearly, 2x2-3 x+6 is a polynomial in one variable because it has only non-negative integral powers of x.
Page No 92:
Question 2:
Which of the following expression is a polynomial?
(a) x-1
(b) x-1x +1
(c) x2-2x2+5
(d) x2+2x3/2x+6
Answer:
(d) x2+2x3/2x+6
We have:
x2+2x32 x+6=x2+2x32x-12+6
=x2+2x+ 6
It a polynomial because it has only non-negative integral powers of x.
Page No 93:
Question 3:
(a) y3+4
(b) y-3
(c) y
(d) 1 y+7
Answer:
(c) y
y is a polynomial because it has a non-negative integral power 1.
Page No 93:
Question 4:
Which of the following is a polynimial?
(a) x-1x+2
(b) 1x+5
(c) x+3
(d) −4
Answer:
(d) −4
-4 is a constant polynomial of degree zero.
Page No 93:
Question 5:
Which of the following is a
polynomial?
(a) x−2 + x−1 + 3
(b) x + x−1 + 2
(c) x−1
(d) 0
Answer:
(d) 0
0 is a polynomial whose degree is not defined.
Page No 93:
Question 6:
Which of the following is quadratic polynomial?
(a) x + 4
(b)x3 + x
(c) x3 + 2x + 6
(d) x2 + 5x + 4
Answer:
(d) x2 + 5x + 4
x2+5x+4 is a polynomial of degree 2. So, it is a quadratic polynomial.
Page No 93:
Question 7:
Which of the
following is a linear polynomial?
(a) x + x2
(b) x + 1
(c) 5x2 − x + 3
(d) x+1x
Answer:
(b) x + 1
Clearly, x+1 is a polynomial of degree 1. So, it is a linear polynomial.
Page No 93:
Question 8:
Which of the following is a binomial?
(a) x2 + x + 3
(b) x2 + 4
(c) 2x2
(d) x+3+1x
Answer:
(b) x2 + 4
Clearly, x2 +4 is an expression having two non-zero terms. So, it is a binomial.
Page No 93:
Question 9:
3is a polynomial of degree
(a) 12
(b) 2
(c) 1
(d) 0
Answer:
(d) 0
3 is a constant term, so it is a polynomial of degree 0.
Page No 93:
Question 10:
Degree of the zero polynomial is
(a) 1
(b) 0
(c) not defined
(d) none of these
Answer:
(c) not defined
Degree of the zero polynomial isnot defined.
Page No 93:
Question 11:
Zero of
the zero polynomial is
(a) 0
(b) 1
(c) every real number
(d) not defined
Answer:
(d) not defined
Zero of the zero polynomial is not defined.
Page No 93:
Question 12:
If p(x) = x = 4, then p(x) + p(−x) = ?
(a) 0
(b) 4
(c) 2x
(d) 8
Answer:
(d) 8
Let:
px=x+4
∴ p-x=-x+4 =-x+4
Thus, we
have:
px+p-x=x+4+ -x+4
= 4 + 4
=8
Page No 93:
Question 13:
If p(x)=x2-22x+1 , then p22=?
(a) 0
(b) 1
(c) 42
(d) −1
Answer:
(b) 1
px=x2-2 2 x+1∴ p22=222-22×22+1
= 8 - 8 + 1
= 1
Page No 93:
Question 14:
If p(x) = 5x– 4x2 + 3 then p(–1) =?
(a) 2
(b) –2
(c) 6
(d) –6
Answer:
p(x) = 5x – 4x2 + 3
Putting x = –1 in p(x), we get
p(–1) = 5 × (–1) – 4 × (–1)2 + 3 = –5 – 4 + 3 = –6
Hence, the correct answer is option (d).
Page No 93:
Question 15:
If (x51 + 51) is divided by (x+ 1) then the remainder is
(a) 0
(b) 1
(c) 49
(d)
50
Answer:
Let f(x) = x51 + 51
By remainder theorem, when f(x) is divided by (x + 1), then the remainder = f(−1).
Putting x = −1 in f(x), we get
f(−1) = (−1)51 + 51 = −1 + 51 = 50
∴ Remainder = 50
Thus, the remainder when (x51 + 51) is divided by (x + 1) is 50.
Hence, the correct answer is option (d).
Page No 93:
Question 16:
If (x + 1) is a
factor of (2x2 + kx), then k = ?
(a) −3
(b) −2
(c) 2
(d) 4
Answer:
(c) 2
x+1 is a factor of 2x2+kx.So, -1 is a zero of 2x2+kx.Thus, we have:2×-12+k×-1=0⇒2-k=0⇒k=2
Page No 93:
Question 17:
When p(x) = x4 + 2x3 − 3x2 + x − 1 is divided by (x − 2), the remainder is
(a) 0
(b) −1
(c) −15
(d) 21
Answer:
(d) 21
x-2=0⇒x=2
By the remainder
theorem, we know that when p(x) is divided by (x - 2), the remainder is p(2).
Thus, we have:
p2=24+2×23-3×22+2-1 =16+16-12+1 =21
Page No 94:
Question 18:
When p(x) = x3 − 3x2 + 4x + 32 is divided by (x + 2), the remainder is
(a) 0
(b) 32
(c) 36
(d) 4
Answer:
(d) 4
x+2=0⇒x=-2
By the remainder theorem, we know that when p(x) is
divided by (x + 2), the remainder is p(-2).
Now, we have:
p-2=-23-3×-22+4×-2+ 32 =-8-12-8+32 =4
Page No 94:
Question 19:
When p(x) = 4x3 − 12x2 + 11x − 5 is divided by (2x − 1), the remainder is
(a) 0
(b) −5
(c) −2
(d) 2
Answer:
(c) −2
2x-1=0⇒x=12
By the remainder theorem, we know that when p(x) is divided by
(2x - 1), the remainder is p12.
Now, we have:
p12=4×1 23-12×122+11×12-5 =12-3+112 -5 =-2
Page No 94:
Question 20:
When p(x) = x3– ax2+ x is divided by (x – a), the remainder is
(a) 0
(b) a
(c) 2a
(d) 3a
Answer:
By remainder theorem, when p(x) = x3 – ax2 + x is divided by (x – a), then the remainder = p(a).
Putting x = a in p(x), we get
p(a) = a3 – a × a2 + a = a3 – a3 + a = a
∴ Remainder = a
Hence, the correct answer is option (b).
Page No 94:
Question 21:
When p(x) = x3 + ax2 + 2x + a is divided by (x + a), the remainder is
(a) 0
(b) −1
(c) −15
(d) 21
Answer:
(c) −a
x+a=0⇒x=-a
By the remainder
theorem, we know that when p (x) is divided by (x + a), the remainder is p (−a).
Thus, we have:
p-a =-a3+a×-a2+2×-a+a =- a3+a3-2a+a =-a
Page No 94:
Question 22:
(x + 1) is a factor of the polynomial
(a) x3 − 2x2 + x + 2
(b) x3+ 2x2 + x − 2
(c) x3 − 2x2 − x − 2
(d) x3 − 2x2 − x + 2
Answer:
(c) x3 − 2x2 − x − 2
Let:
f x=x3-2x2+x+2
By the factor theorem, (x + 1) will be a factor of f (x) if f (−1) = 0.
We have:
f-1=-13 -2×-12+-1+2 =-1-2-1+2 =-2≠0
Hence, (x + 1) is not a factor of fx=x3-2x2+x+2.
Now,
Let:
fx=x3+2x2+x-2
By the factor theorem, (x + 1) will be a factor of f (x) if f (-1)
= 0.
We have:
f -1=-13+2×-12+-1-2 =- 1+2-1-2 =-2≠0
Hence, (x + 1) is not a factor of fx=x3+ 2x2+x-2.
Now,
Let:
fx=x3+2x2-x-2
By the factor theorem, (x + 1) will be a factor of f (x) if f (-1) = 0.
We have:
f-1=-13+2×-12--1-2 =-1+2+1-2 =0
Hence, (x + 1) is a factor of fx=x3+2x2-x-2.
Page No 94:
Question 23:
Zero of the polynomial p(x) = 2x + 5 is
(a) -25
(b) -52
(c) 25
(d) 52
Answer:
The zero of the polynomial p(x) can be obtained by putting p(x) = 0.
px=0⇒2x+5=0⇒2x=-5⇒x=-52
Hence, the correct answer is option (b).
Page No 94:
Question 24:
The zeros of the polynomial p(x) = x2 + x – 6 are
(a) 2, 3
(b) –2, 3
(c) 2, –3
(d) –2, –3
Answer:
The given polynomial is p(x) = x2 + x – 6.
Putting x = 2 in p(x), we get
p(2) = 22 + 2 – 6 = 4 + 2 – 6 = 0
Therefore, x = 2 is a zero of the polynomial p(x).
Putting x = –3 in p(x), we get
p(–3) = (–3)2 – 3 – 6 = 9 – 9 = 0
Therefore, x = –3 is a zero of the polynomial p(x).
Thus, 2 and –3 are the zeroes of the given polynomial p(x).
Hence, the correct answer is option (c).
Page No 94:
Question 25:
The zeros of the polynomial p(x) = 2x2 + 5x – 3 are
(a) 12, 3
(b) 12, -3
(c) -12, 3
(d) 1,-12
Answer:
The given polynomial is p(x) = 2x2 + 5x – 3.
Putting x=12 in p(x), we get
p12=2×122+5×12-3=12 +52-3=3-3=0
Therefore, x=12 is a zero of the polynomial p(x).
Putting x = –3 in p(x), we get
p -3=2×-32+5×-3-3=18-15-3=0
Therefore, x = –3 is a zero of the polynomial p(x).
Thus, 12 and –3 are the zeroes of the given polynomial p(x).
Hence, the correct answer is option (b).
Page No 94:
Question 26:
The zeros of the polynomial p(x) = 2x2 + 7x – 4 are
(a) 4, -12
(b) 4, 12
(c) -4, 12
(d) -4, -12
Answer:
The given polynomial is p(x) = 2x2 + 7x – 4.
Putting x=12 in p(x), we get
p12=2× 122+7×12-4=12+72-4=4-4=0
Therefore, x=12 is a zero of the polynomial p(x).
Putting x = –4 in p(x), we get
p-4=2×-42+7×-4-4=32-28-4=32- 32=0
Therefore, x = –4 is a zero of the polynomial p(x).
Thus, 12 and –4 are the zeroes of the given polynomial p(x).
Hence, the correct answer is option (c).
Page No 94:
Question 27:
If (x + 5) is a factor of p(x) = x3 − 20x + 5k, then k = ?
(a) −5
(b) 5
(c) 3
(d) −3
Answer:
(b) 5
x+5 is a factor of px=x3-20x+5k .∴ p-5=0⇒-53-20×-5+5k=0⇒-125+100 +5k=0⇒5k=25⇒k=5
Page No 94:
Question 28:
If (x + 2) and (x − 1) are factors of (x3
+ 10x2 + mx + n), then
(a) m = 5, n = −3
(b) m = 7, n = −18
(c) m = 17, n = −8
(d) m = 23, n = −19
Answer:
(b) m = 7, n = −18
Let:
px=x3+10x2+mx+n
Now,
x+2=0⇒x=-2
(x + 2) is a factor of p(x).
So, we have p(-2)=0
⇒-23 +10×-22+m×-2+n=0⇒-8+40-2m+n=0⇒32-2m+n =0⇒2m-n=32 .....i
Now,
x-1=0⇒x=1
Also,
(x - 1) is a factor of p(x).
We have:
p(1) = 0
⇒13+10×12+m×1+n=0⇒1+10+m+n=0⇒11+m+n=0 ⇒m+n=-11 .....iiFrom i and ii, we get:3m=21⇒m =7
By substituting the value of m in (i), we get n = −18.
∴ m = 7 and n = −18
Page No 94:
Question 29:
If (x100 + 2x99 + k) is divisible by (x + 1), then the value of k is
(a) 1
(b) 2
(c) −2
(d) −3
Answer:
(a) 1
Let:
px=x100+2x99+k
Now,
x+1=0⇒x=-1
x+1 is divisible by px.∴p-1=0⇒-1100+2×-199+k=0 ⇒1-2+k=0⇒-1+k=0⇒k=1
Page No 94:
Question 30:
For what value of k is the polynomial p(x) = 2x3 − kx + 3x + 10 exactly divisible by (x + 2)?
(a) -13
(b) 13
(c) 3
(d)
−3
Answer:
(d) −3
Let:
px=2x3- kx2+3x+10
Now,
x+2=0⇒x=-2
px is completely divisible by x +2.∴ p-2=0⇒2×-23-k×-22+3×- 2+10=0⇒-16-4k-6+10=0⇒-12-4k=0⇒4k=-12⇒k =-124⇒k=-3
Page No 94:
Question 31:
The zeroes of the polynomial p(x) = x2 − 3x are
(a) 0, 0
(b) 0, 3
(c) 0, −3
(d) 3, −3
Answer:
(b) 0, 3
Let:
px=x2-3x
Now, we have:
px=0⇒x2-3x=0
⇒xx-3 =0⇒x=0 and x-3=0⇒x=0 and x=3
Page No 95:
Question 32:
The zeroes of the polynomial p(x) = 3x2 − 1 are
(a) 13
(b) 13
(c) -13
(d) 13and -13
Answer:
(d) 13 and -13
Let:
px=3x 2-1
To find the zeroes of px, we have:px=0⇒ 3x2-1=0
⇒3x2=1⇒x2=13⇒x=±13 ⇒x=13 and x=-13
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