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Page No 74:Question 1:Which of the following expressions are polynomials? In case of a polynomial, write its degree. 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 (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 (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 (xv) 2x3+3x2+x-1=2x3+3x2+
x12-1 Page No 75:Question 2:Identify constant, linear, quadratic, cubic and quartic polynomials from the following. 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 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 (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. Answer: (i) 4x-5x2 +6x32x=4x2x-5x22x+6x32x=2-52x +3x2 (ii) y2(y – y3) = y3 – y5 (iii) (3x – 2)(2x3 + 3x2) = 6x4 + 9x3 – 4x3 – 6x2 = 6x4 + 5x3 – 6x2 (iv) -12x+3 (v) – 8 (vi) x–2(x4 + x2) = x2 + x0 = x2 + 1 Page No 75:Question 5:(i) Give an example of a monomial of degree 5. 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. (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. (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. (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. Page No 75:Question 6:Rewrite each of the following polynomials in standard form. 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. 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 ii px=5-4x+2x2⇒p3=5-4×3+2×32 iii px=5-4x+2x2⇒p-2=5-4×-2+2×-22 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 ii py= 4+3y-y2+5y3⇒p2=4+3×2-22+5×23 iii py=4+3y-y2+5y3 ⇒p-1=4+3×-1--12+5×-13 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 ii ft=4t2-3t+6⇒f4=4×42-3×4+6 iii ft=4t2-3t+6⇒f-5=4×-52-3 ×-5+6 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: Answer:i px=x-4⇒p 4=4-4 ii px=-3+3⇒p 3=0 iii px=2-5x⇒p25=2-5×25 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 Answer:i px=x2-3x+2=x-1x-2⇒p1=1-1×1-2 Hence, 1 and 2 are the zeroes of the given polynomial. ii px=x2+x-6⇒p2=22+2-6 Hence, 2 and -3 are the zeroes of the given polynomial. iii px=x2-3x⇒p0=02-3×0 Also, Hence, 0 and 3 are the zeroes of the given polynomial. Page No 79:Question 8:Find the zero of the polynomial: 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. Answer:
Page No 84:Question 1:By actual division, find the quotient and the remainder when (x4 + 1) is divided by (x – 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 Hence verified. Page No 84:Question 3:Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where 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 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 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 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 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 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 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 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 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 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 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. Now, 2-a+b=5 ...1 6+a+b=19 ...2 Adding (1) and (2), we get:8+2b=24 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: Answer:Let: Page No 90:Question 2:Using factor theorem, show
that: Answer:Let: Page No 90:Question 3:Using factor theorem, show that: Answer:Let: Page No 91:Question 4:Using factor theorem, show
that: Answer:Let: Page No 91:Question 5:Using factor theorem,
show that g(x) is a factor of p(x), when 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: Answer:Let: Page No 91:Question 7:Using factor theorem, show that: Answer:Let: f32=2× 324+323-8×322-32+6 =818+278-18-32+6 =0 Page No 91:Question 8:Using factor
theorem, show that g(x) is a factor of p(x), when 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: Answer:Let: Page No 91:Question 10:Using factor theorem, show that: Answer:Let: 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: Page No 91:Question 13:Find the value of a for which (x − 4) is a factor of (2x3 − 3x2 − 18x + a). Answer:Let: 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 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 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 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: Page No 91:Question 18:Without actual division, show that (x3 − 3x2 − 13x + 15) is exactly divisible by (x2 + 2x − 3). Answer:Let: 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: Now, 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: Subtracting (1) from (2), we get:a=23 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: 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 Using factor theorem, we have f12=0⇒p×122+5×12+r=0 ⇒p4+r=-52⇒p+4r=-10 .....2 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 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 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? 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? Answer:(d) x2+2x3/2x+6 We have: x2+2x32 x+6=x2+2x32x-12+6 It a polynomial because it has only non-negative integral powers of x. Page No 93:Question 3:(a) y3+4 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? Answer:(d) −4 -4 is a constant polynomial of degree zero. Page No 93:Question 5:Which of the following is a
polynomial? Answer:(d) 0 0 is a polynomial whose degree is not defined. Page No 93:Question 6:Which of the following is quadratic polynomial? 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? 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? 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 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 Answer:(c) not defined Degree of the zero polynomial isnot defined. Page No 93:Question 11:Zero of
the zero polynomial is 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) = ? Answer:(d) 8 Let: ∴ p-x=-x+4 =-x+4 Page No 93:Question 13:If p(x)=x2-22x+1 , then p22=? Answer:(b) 1 px=x2-2 2 x+1∴ p22=222-22×22+1 Page No 93:Question 14:If p(x) = 5x– 4x2 + 3 then p(–1) =? 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 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 = ? 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 Answer:(d) 21 x-2=0⇒x=2 Page No 94:Question 18:When p(x) = x3 − 3x2 + 4x + 32 is divided by (x + 2), the remainder is Answer:(d) 4 x+2=0⇒x=-2 Page No 94:Question 19:When p(x) = 4x3 − 12x2 + 11x − 5 is divided by (2x − 1), the remainder is Answer:(c) −2 2x-1=0⇒x=12 Page No 94:Question 20:When p(x) = x3– ax2+ x is divided by (x – a), the remainder is 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 Answer:(c) −a x+a=0⇒x=-a Page No 94:Question 22:(x + 1) is a factor of the polynomial Answer:(c) x3 − 2x2 − x − 2 Let: Now, Now, 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 Page No 94:Question 24:The zeros of the polynomial p(x) = x2 + x – 6 are 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 = ? 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 Answer:(b) m = 7, n = −18 Let: Page No 94:Question 29:If (x100 + 2x99 + k) is divisible by (x + 1), then the value of k is Answer:(a) 1 Let: Page No 94:Question 30:For what value of k is the polynomial p(x) = 2x3 − kx + 3x + 10 exactly divisible by (x + 2)? Answer:(d) −3 Let: Page No 94:Question 31:The zeroes of the polynomial p(x) = x2 − 3x are Answer:(b) 0, 3 Let: Page No 95:Question 32:The zeroes of the polynomial p(x) = 3x2 − 1 are Answer:(d) 13 and -13 Let: To find the zeroes of px, we have:px=0⇒ 3x2-1=0 View NCERT Solutions for all chapters of Class 6 What must be added to x³ 3x² 12x 19 so that the result is exactly divisible by X² x 6?Solution : Hence, `-(2x+5)` should be added to `x^3-3x^2-12 x+19` so that the result is exactly divisible by `x^2+x-6`.
What must be added to obtain a polynomial which is exactly divisible by x 3?<br> Hence , the required number to be added is `3` .
What must be added to X cube minus 3 x Square 4 x 13 to obtain polynomial exactly divisible by X minus 3?∴{(3)3−3(3)2+4(3)−13}+Y=0 . (x3−3x2+4x−13) to obtain a polynomial which is exactly divisible by.. x−3. . What should be added to the polynomial x3 3x2 6x 15 so that it is completely divisible by x 3?On dividing x3 – 3x2 + 6x – 15 by x – 3, remainder is + 3, hence – 3 must be added to x3 – 3x2 + 6x – 15.
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