993 + Interview Questions in POLYNOMIALS IN GENERAL AWARENESS Page 4 InterviewSolution

151.	1012 = ...................A) 12001B) 10001 C) 10201 D) 100201
Answer» Correct option is (C) 10201 \(101^2=(100+1)^2\) \(=100^2+2\times100\times1+1^2\) = 10000+200+1 = 10201 Correct option is C) 10201

151.

1012 = ...................A) 12001B) 10001 C) 10201 D) 100201

Answer»

Correct option is (C) 10201

\(101^2=(100+1)^2\) \(=100^2+2\times100\times1+1^2\)

= 10000+200+1 = 10201

Correct option is C) 10201

Discussion

152.	Which of the following expressions is not a polynomial ?A) \(\frac{1}{x\, + 1}\)B) 7 C) x2 – 3 D) x2 – 4x + 5
Answer» Correct option is (A) \(\frac{1}{x+1}\) \(\frac{1}{x+1}\) is a fraction (rational expression) So, \(\frac{1}{x+1}\) is not a polynomial. Correct option is A) \(\frac{1}{x\, + 1}\)

152.

Which of the following expressions is not a polynomial ?A) \(\frac{1}{x\, + 1}\)B) 7 C) x2 – 3 D) x2 – 4x + 5

Answer»

Correct option is (A) \(\frac{1}{x+1}\)

\(\frac{1}{x+1}\) is a fraction (rational expression)

So, \(\frac{1}{x+1}\) is not a polynomial.

Correct option is A) \(\frac{1}{x\, + 1}\)

Discussion

153.	Which of the following is NOT a polynomial ? A) 5 B) \(\sqrt3x^2\) + 5y C) \(\sqrt{x}\)D) 3xyz
Answer» Correct option is (C) \(\sqrt{x}\) \(\sqrt{x}\) has a rational power. \(\therefore\sqrt{x}\) is not a polynomial. Correct option is C) \(\sqrt{x}\)

153.

Which of the following is NOT a polynomial ? A) 5 B) \(\sqrt3x^2\) + 5y C) \(\sqrt{x}\)D) 3xyz

Answer»

Correct option is (C) \(\sqrt{x}\)

\(\sqrt{x}\) has a rational power.

\(\therefore\sqrt{x}\) is not a polynomial.

Correct option is C) \(\sqrt{x}\)

Discussion

154.	A quadratic polynomial in one variable has………….terms. A) 1 B) 2 C) 3 D) 4
Answer» Correct option is (C) 3 Quadratic polynomial in one variable is \(ax^2+bx+c\) which has three terms. Thus, a quadratic polynomial in one variable has 3 terms. Correct option is C) 3

154.

A quadratic polynomial in one variable has………….terms. A) 1 B) 2 C) 3 D) 4

Answer»

Correct option is (C) 3

Quadratic polynomial in one variable is \(ax^2+bx+c\) which has three terms.

Thus, a quadratic polynomial in one variable has 3 terms.

Correct option is C) 3

Discussion

155.	If p(x) = x2 + 5x + 4, then p(- 4) = A) 0 B) – 32 C) 40 D) – 20
Answer» Correct option is (A) 0 \(\because\) \(p(x)=x^2+5x+4\) \(\therefore\) \(p(-4)=(-4)^2+5\times-4+4\) = 16 - 20 + 4 = 20 - 20 = 0 Correct option is A) 0

155.

If p(x) = x2 + 5x + 4, then p(- 4) = A) 0 B) – 32 C) 40 D) – 20

Answer»

Correct option is (A) 0

\(\because\) \(p(x)=x^2+5x+4\)

\(\therefore\) \(p(-4)=(-4)^2+5\times-4+4\)

= 16 - 20 + 4 = 20 - 20 = 0

Correct option is A) 0

Discussion

156.	The first term in quotient when you divide 3x2 + x – 1 by x+1 is A) 1 B) – 2 C) 3x D) 3
Answer» Correct option is (C) 3x \(\frac{3x^2+x-1}{x+1}=3x-2+\frac1{x+1}\) \(\therefore\) Quotient = 3x - 2 First term in quotient = 3x Correct option is C) 3x

156.

The first term in quotient when you divide 3x2 + x – 1 by x+1 is A) 1 B) – 2 C) 3x D) 3

Answer»

Correct option is (C) 3x

\(\frac{3x^2+x-1}{x+1}=3x-2+\frac1{x+1}\)

\(\therefore\) Quotient = 3x - 2

First term in quotient = 3x

Correct option is C) 3x

Discussion

157.	Find the value of the polynomial `5x-4x^(2)+3` at : `(i)x=0 " " (ii) x=-1 " " (iii) x=2`
Answer» Let `p(x)=5x-4x^(2)+3` (i) The value of `p(x)=5x-4x^(2)+3 " at" x=0 " is"` `p(0)=5xx0-4xx0^(2)+3` `implies p(0)=3` (ii) The value of `p(x)=5x-4x^(2)+3 " at" x=-1 ` is `p(-1)=5(-1)-4(-1)^(2)+3=-5-4+3` `implies p(-1)=-6` (iii) The value of `p(x)=5x-4x^(2)+3 " at" x=2` is `p(2)=5(2)-4(2)^(2)+3=10-16+3` `implies p(2)=-3`

Discussion

158.	The identity used in simplifying 103 x 97 A) (a + b)2 = a2 + 2ab + b2 B) (a – b)2 = a2 – 2ab + b2 C) (a – b) (a + b) = a2 – b2 D) (x + a) (x + b) = x2 + (a + b) x + ab
Answer» Correct option is (C) (a – b) (a + b) = a² – b² \(103\times97\) = (100 + 3) (100 - 3) Let a = 100, b = 3 Then \(103\times97\) \(=(a+b)(a-b)\) \(=a^2-b^2\) \(=100^2-3^2\) = 10000 - 9 = 9991 \(\therefore\) Identity that is used in simplifying \(103\times97\) is \((a-b)(a+b)=a^2-b^2.\) C) (a – b) (a + b) = a² – b²

158.

The identity used in simplifying 103 x 97 A) (a + b)2 = a2 + 2ab + b2 B) (a – b)2 = a2 – 2ab + b2 C) (a – b) (a + b) = a2 – b2 D) (x + a) (x + b) = x2 + (a + b) x + ab

Answer»

Correct option is (C) (a – b) (a + b) = a² – b²

\(103\times97\) = (100 + 3) (100 - 3)

Let a = 100, b = 3

Then \(103\times97\) \(=(a+b)(a-b)\) \(=a^2-b^2\)

\(=100^2-3^2\)

= 10000 - 9 = 9991

\(\therefore\) Identity that is used in simplifying \(103\times97\) is \((a-b)(a+b)=a^2-b^2.\)

C) (a – b) (a + b) = a² – b²

Discussion

159.	99 x 99 .......................A) II B) III C) I and II D) I
Answer» Correct option is D) I

Discussion

160.	Which of the following expressions is a polynomial ?A) 3x-2 + 7x + 1B) 7x2 + 2/x – 8C) x2 – x – 1D) x +x/2 + 1/x
Answer» Correct option is (C) x² – x – 1 \(\because\) x - 2, \(\frac2x\) & \(\frac1x\) are rational expression (fractions) and polynomial are free from fractions. \(\therefore\) These are not an example of polynomials. Also, \(x^2-x-1\) is a polynomial because variables has natural powers. C) x² – x – 1

160.

Which of the following expressions is a polynomial ?A) 3x-2 + 7x + 1B) 7x2 + 2/x – 8C) x2 – x – 1D) x +x/2 + 1/x

Answer»

Correct option is (C) x² – x – 1

\(\because\) x - 2, \(\frac2x\) & \(\frac1x\) are rational expression (fractions) and polynomial are free from fractions.

\(\therefore\) These are not an example of polynomials.

Also, \(x^2-x-1\) is a polynomial because variables has natural powers.

C) x² – x – 1

Discussion

161.	If 2x3 – 2x2 – 2x – 5 is divided by x + 1, then the remainder is A) 0B) -7 C) 6 D) -6
Answer» Correct option is (B) -7 Let \(p(x)=2x^3-2x^2-2x-5\) When p(x) is divided by (x+1), then it leaves remainder p(-1). Now, p(-1) \(=2(-1)^3-2(-1)^2-2\times-1-5\) = -2 - 2 + 2 - 5 = -7 \(\therefore\) If \(2x^3-2x^2-2x-5\) is divided by (x+1), then the leaving remainder is -7. Correct option is B) -7

161.

If 2x3 – 2x2 – 2x – 5 is divided by x + 1, then the remainder is A) 0B) -7 C) 6 D) -6

Answer»

Correct option is (B) -7

Let \(p(x)=2x^3-2x^2-2x-5\)

When p(x) is divided by (x+1), then it leaves remainder p(-1).

Now, p(-1) \(=2(-1)^3-2(-1)^2-2\times-1-5\)

= -2 - 2 + 2 - 5 = -7

\(\therefore\) If \(2x^3-2x^2-2x-5\) is divided by (x+1), then the leaving remainder is -7.

Correct option is B) -7

Discussion

162.	The remainder when 2x3 + 3x2 +4x + 5 is divided by (x – 2) A) 31B) 29 C) 36 D) 41
Answer» Correct option is (D) 41 Let p(x) = \(2x^3+3x^2+4x+5\) When p(x) is divided by \((x-2)\) then remainder will be p(2). \(\therefore\) Remainder = p(2) \(=2\times2^3+3\times2^2+4\times2+5\) = 16+12+8+5 = 41 Correct option is D) 41

162.

The remainder when 2x3 + 3x2 +4x + 5 is divided by (x – 2) A) 31B) 29 C) 36 D) 41

Answer»

Correct option is (D) 41

Let p(x) = \(2x^3+3x^2+4x+5\)

When p(x) is divided by \((x-2)\) then remainder will be p(2).

\(\therefore\) Remainder = p(2) \(=2\times2^3+3\times2^2+4\times2+5\)

= 16+12+8+5 = 41

Correct option is D) 41

Discussion

163.	The remainder when x3 + 2x2 + 3x + 4 is divided by (x – 1) is A) 10 B) 6 C) 4D) 2
Answer» Correct option is (A) 10 Let p(x) = \(x^3+2x^2+3x+4\) When p(x) is divided by \((x-1)\) then remainder will be p(1). \(\therefore\) Remainder = p(1) \(=1^3+2\times1^2+3\times1+4\) = 1+2+3+4 = 10 Correct option is A) 10

163.

The remainder when x3 + 2x2 + 3x + 4 is divided by (x – 1) is A) 10 B) 6 C) 4D) 2

Answer»

Correct option is (A) 10

Let p(x) = \(x^3+2x^2+3x+4\)

When p(x) is divided by \((x-1)\) then remainder will be p(1).

\(\therefore\) Remainder = p(1) \(=1^3+2\times1^2+3\times1+4\)

= 1+2+3+4 = 10

Correct option is A) 10

Discussion

164.	The remainder when x101 + 101 is divided by x + 1 is A) 1 B) 100 C) 101 D) 0
Answer» Correct option is (B) 100 Let p(x) = \(x^{101}+101\) \(\because\) When p(x) is divided by (x+1), it leaves remainder p(-1). Now, p(-1) = \((-1)^{101}+101\) = -1+101 = 100 Hence, the remainder when \(x^{101}+101\) is divided by \((x+1)\) is 100. Correct option is A) 1

164.

The remainder when x101 + 101 is divided by x + 1 is A) 1 B) 100 C) 101 D) 0

Answer»

Correct option is (B) 100

Let p(x) = \(x^{101}+101\)

\(\because\) When p(x) is divided by (x+1), it leaves remainder p(-1).

Now, p(-1) = \((-1)^{101}+101\) = -1+101 = 100

Hence, the remainder when \(x^{101}+101\) is divided by \((x+1)\) is 100.

Correct option is A) 1

Discussion

165.	Zero of the polynomial p(x) = 3x + 1 is A) 1/3B) – 2 C) -1/3D) 3
Answer» Correct option is (C) -1/3 Given polynomial is \(p(x)=3x+1\) p(x) = 0 gives 3x+1 = 0 \(\Rightarrow\) \(x=\frac{-1}3\) Hence, \(\frac{-1}3\) is a zero of polynomial \(3x+1.\) Correct option is C) -1/3

165.

Zero of the polynomial p(x) = 3x + 1 is A) 1/3B) – 2 C) -1/3D) 3

Answer»

Correct option is (C) -1/3

Given polynomial is \(p(x)=3x+1\)

p(x) = 0 gives 3x+1 = 0

\(\Rightarrow\) \(x=\frac{-1}3\)

Hence, \(\frac{-1}3\) is a zero of polynomial \(3x+1.\)

Correct option is C) -1/3

Discussion

166.	Find the square root of `(2x+7)(x^(2)-9)(2x-5)+9`.
Answer» First of all we make the coefficient of x as 1 in each factor by taking some constant common as : `2(x+(7)/(2))(x+3)(x-3).2(x-(5)/(2))+9` Now, we take the product of 2 factors at a time in such a way that sum of constant terms is same in each pair as : ` (##NTN_MATH_IX_C02_S01_066_S01.png" width="80%"> `=4(x^(2)+(x)/(2)-(21)/(2))(x^(2)+(x)/(2)-(15)/(2))+9` Let `x^(2)+(x)/(2)=a` `=(a-(21)/(2))(a-(15)/(2))+9` `4(a^(2)-18a+(315)/(4))+9` `4(a^(2)-18a)+315+9` `4underset("make perfect square")(ubrace(a^(2)-18a+81)-81)+324 " " ["add and subtract"(("coeff. of a")/(2))^(2) " to make perfect square"]` `=4(a^(2)-18a+81)-324+324` `=4(a-9)^(2)` Replace the value of a as `x^(2)+(x)/(2)`, we get `4(x^(2)+(x)/(2)-9)^(2)=4((2x^(2)+x-18)/(2))^(2)=(2x^(2)+x-18)^(2)` `therefore " Square root of" (2x+7)(x^(2)-9)(2x-5)+9 i.e., (2x^(2)+x-18)^(2) " is" 2x^(2)+x-18`.

Discussion

167.	Which of the following figure represent the algebrical identity (x – y)2 = x2 – 2xy + y2 ?
Answer» Correct option is (B)

Discussion

168.	If `abx^(2)=(a-b)^(2)(x+1)`, then find the value of `1+(4)/(x)+(4)/(x^(2))` in terms of a and b.
Answer» Since `abx=(a-b)^(2)(x+1)` `implies (x+1)/(x^(2))=(ab)/((a-b)^(2)) implies (1)/(x)+(1)/(x^(2))=(Ab)/((a-b)^(2))` `implies (4)/(x)+(4)/(x^(2))=(4ab)/((a-b)^(2))` `implies 1+(4)/(x)+(4)/(x^(2))=1+(4ab)/((a-b)^(2))` `=(a^(2)+b^(2)-2ab+4ab)/((a-b)^(2))=(a^(2)+b^(2)+2ab)/((a-b)^(2))=((a+b)^(2))/((a-b)^(2))`

Discussion

169.	The value of p(x) = 7x2 + 2x – 8 when x = 1 A) – 1 B) 1 C) 2 D) 17
Answer» Correct option is (B) 1 \(\because\) \(p(x)\) = \(7x^2+2x-8\) \(\therefore\) \(p(1)\) = \(7.1^2+2.1-8\) = 7+2-8 = 9 - 8 = 1 Correct option is B) 1

169.

The value of p(x) = 7x2 + 2x – 8 when x = 1 A) – 1 B) 1 C) 2 D) 17

Answer»

Correct option is (B) 1

\(\because\) \(p(x)\) = \(7x^2+2x-8\)

\(\therefore\) \(p(1)\) = \(7.1^2+2.1-8\) = 7+2-8

= 9 - 8 = 1

Correct option is B) 1

Discussion

170.	The zero of the polynomial px + q is… A) p/qB) q/pC) -(p/q)D) -(q/p)
Answer» Correct option is (D) -(q/p) \(\because\) \(px+q\) = 0 \(\Rightarrow\) x = \(\frac{-q}{p}\) \(\therefore\) \(\frac{-q}{p}\) is a zero of polynomial \(px+q.\) Correct option is D) -(q/p)

170.

The zero of the polynomial px + q is… A) p/qB) q/pC) -(p/q)D) -(q/p)

Answer»

Correct option is (D) -(q/p)

\(\because\) \(px+q\) = 0

\(\Rightarrow\) x = \(\frac{-q}{p}\)

\(\therefore\) \(\frac{-q}{p}\) is a zero of polynomial \(px+q.\)

Correct option is D) -(q/p)

Discussion

171.	The remainder when the polynomial f(x) in x is divided by 3x + 2 is ……………..A) f(2) B) f(-2) C) f(2/3 ) D) f(\(-\frac{2}{3}\))
Answer» Correct option is (D) f\((-\frac{2}{3})\) The remainder when the polynomial \(f(x)\) is divided by \(3x+2\) is \(f(-\frac{2}{3}).\) \((\because3x+2=0\Rightarrow x=\frac{-2}3)\) Correct option is D) f(\(-\frac{2}{3}\))

171.

The remainder when the polynomial f(x) in x is divided by 3x + 2 is ……………..A) f(2) B) f(-2) C) f(2/3 ) D) f(\(-\frac{2}{3}\))

Answer»

Correct option is (D) f\((-\frac{2}{3})\)

The remainder when the polynomial \(f(x)\) is divided by \(3x+2\) is \(f(-\frac{2}{3}).\) \((\because3x+2=0\Rightarrow x=\frac{-2}3)\)

Correct option is D) f(\(-\frac{2}{3}\))

Discussion

172.	If x + 1 is a factor of polynomial 2x2 + Kx then the value of K isA) -4 B) -2 C) 2 D) +4
Answer» Correct option is (C) 2 \(\because\) \(x+1\) is a factor of polynomial \(2x^2+Kx.\) \(\therefore\) x = -1 is a zero of polynomial \(2x^2+Kx.\) \(\therefore\) \(2\times(-1)^2+K\times-1=0\) \(\Rightarrow\) K = 2 Correct option is C) 2

172.

If x + 1 is a factor of polynomial 2x2 + Kx then the value of K isA) -4 B) -2 C) 2 D) +4

Answer»

Correct option is (C) 2

\(\because\) \(x+1\) is a factor of polynomial \(2x^2+Kx.\)

\(\therefore\) x = -1 is a zero of polynomial \(2x^2+Kx.\)

\(\therefore\) \(2\times(-1)^2+K\times-1=0\)

\(\Rightarrow\) K = 2

Correct option is C) 2

Discussion

173.	One of the factors of `(25 x^(2)-1) +(1+5x)^(2)` isA. `5+x`B. `5-x`C. `5x-1`D. `10x`
Answer» Correct Answer - D Now ,`(25x^(2)-1)=(1+5x)^(2)` `=25x^(2)-1+1+25x^(2)+10x " " ["using identity , " (a+b)^(2)=a^(2)+b^(2)+2ab]` `=50 x^(2)+10x=10x(5x+1)` Hence ,one of the factor of given polynomial is 10x .

Discussion

174.	If the area of a square field is x2 + y2 + z2 + 2xy – 2yz – 2zx sq. units, then its perimeter is A) (x + y + z)2 B) 4 (x + y + z) C) 4 (x + y – z) D) (x + y – z)2
Answer» Correct option is (C) 4 (x + y – z) Let side length of the square be a unit. \(\therefore a^2=x^2+y^2+z^2+2xy-2yz-2zx\) \(=x^2+y^2+(-z)^2+2\times x\times y\) \(+2\times y\times -z+2\times -z\times x\) \(=(x+y-z)^2\) sq. units \(\Rightarrow\) \(a=(x+y-z)\) unit \(\therefore\) Perimeter of the square = 4a = \(4\,(x+y-z)\) unit C) 4 (x + y – z)

174.

If the area of a square field is x2 + y2 + z2 + 2xy – 2yz – 2zx sq. units, then its perimeter is A) (x + y + z)2 B) 4 (x + y + z) C) 4 (x + y – z) D) (x + y – z)2

Answer»

Correct option is (C) 4 (x + y – z)

Let side length of the square be a unit.

\(\therefore a^2=x^2+y^2+z^2+2xy-2yz-2zx\)

\(=x^2+y^2+(-z)^2+2\times x\times y\) \(+2\times y\times -z+2\times -z\times x\)

\(=(x+y-z)^2\) sq. units

\(\Rightarrow\) \(a=(x+y-z)\) unit

\(\therefore\) Perimeter of the square = 4a

= \(4\,(x+y-z)\) unit

C) 4 (x + y – z)

Discussion

175.	The coefficient of x2 in (2x – 8) (7 – 3x) is A) 2 B) 3 C) 6 D) – 6
Answer» Correct option is (D) –6 \(\because\) \((2x-8)\) \((7-3x)\) \(=-6x^2+38x-56\) \(\therefore\) Coefficient of \(x^2\) is -6. Correct option is D) – 6

175.

The coefficient of x2 in (2x – 8) (7 – 3x) is A) 2 B) 3 C) 6 D) – 6

Answer»

Correct option is (D) –6

\(\because\) \((2x-8)\) \((7-3x)\) \(=-6x^2+38x-56\)

\(\therefore\) Coefficient of \(x^2\) is -6.

Correct option is D) – 6

Discussion

176.	The value of p if px4 + 7x2 – 18 is divisible by (x – 3)A) 9/5B) -5/9C) 5/9D) 1
Answer» Correct option is (B) -5/9 \(\because\) \(px^4+7x^2-18\) is divisible by (x - 3). \(\therefore\) When \((px^4-7x^2-18)\) is divided by (x - 3), it leaves remainder 0. \(\therefore\) \(p(3)^4+7\times3^2-18=0\) \(\Rightarrow\) 81p + 63 - 18 = 0 \(\Rightarrow\) 81p = -63 + 18 \(\Rightarrow\) 81p = -45 \(\Rightarrow\) \(p=\frac{-45}{81}=\frac{-5}{9}\) Correct option is B) -5/9

176.

The value of p if px4 + 7x2 – 18 is divisible by (x – 3)A) 9/5B) -5/9C) 5/9D) 1

Answer»

Correct option is (B) -5/9

\(\because\) \(px^4+7x^2-18\) is divisible by (x - 3).

\(\therefore\) When \((px^4-7x^2-18)\) is divided by (x - 3), it leaves remainder 0.

\(\therefore\) \(p(3)^4+7\times3^2-18=0\)

\(\Rightarrow\) 81p + 63 - 18 = 0

\(\Rightarrow\) 81p = -63 + 18

\(\Rightarrow\) 81p = -45

\(\Rightarrow\) \(p=\frac{-45}{81}=\frac{-5}{9}\)

Correct option is B) -5/9

Discussion

177.	If 2 is a zero of the polynomial x2 – kx + 8, then k = A) 1 B) 4 C) 6 D) 3
Answer» Correct option is (C) 6 Given that 2 is a zero of polynomial \(x^2-kx+8.\) \(\therefore2^2-2k+8=0\) \(\Rightarrow\) 2k = 4+8 = 12 \(\Rightarrow\) \(k=\frac{12}2=6\) Correct option is C) 6

177.

If 2 is a zero of the polynomial x2 – kx + 8, then k = A) 1 B) 4 C) 6 D) 3

Answer»

Correct option is (C) 6

Given that 2 is a zero of polynomial \(x^2-kx+8.\)

\(\therefore2^2-2k+8=0\)

\(\Rightarrow\) 2k = 4+8 = 12

\(\Rightarrow\) \(k=\frac{12}2=6\)

Correct option is C) 6

Discussion

178.	Which of the following is not a zero of the polynomial x3 – 6x2 + 11x – 6? A) 1 B) 2 C) 3 D) 0
Answer» Correct option is (D) 0 Let \(p(x)=x^3 – 6x^2 + 11x – 6\) \(\because\) p(0) = -6 \(\neq\) 0 \(\therefore\) 0 is not a zero of given polynomial p(x). Alternative :- \(p(x)=x^3 – 6x^2 + 11x – 6\) \(=x^3-x^2-5x^2+5x+6x-6\) \(=x^2(x-1)-5x(x-1)+6(x-1)\) \(=(x-1)(x^2-5x+6)\) \(=(x-1)(x-2)(x-3)\) \(\therefore\) 1, 2 & 3 are zeros of given polynomial p(x). Correct option is D) 0

178.

Which of the following is not a zero of the polynomial x3 – 6x2 + 11x – 6? A) 1 B) 2 C) 3 D) 0

Answer»

Correct option is (D) 0

Let \(p(x)=x^3 – 6x^2 + 11x – 6\)

\(\because\) p(0) = -6 \(\neq\) 0

\(\therefore\) 0 is not a zero of given polynomial p(x).

Alternative :-

\(p(x)=x^3 – 6x^2 + 11x – 6\)

\(=x^3-x^2-5x^2+5x+6x-6\)

\(=x^2(x-1)-5x(x-1)+6(x-1)\)

\(=(x-1)(x^2-5x+6)\)

\(=(x-1)(x-2)(x-3)\)

\(\therefore\) 1, 2 & 3 are zeros of given polynomial p(x).

Correct option is D) 0

Discussion

179.	Using factor theorem; factorize the polynomial `x^3-6x^2+11x-6`
Answer» Correct Answer - `(i) (x-1)(x-2)(x-3) " " (ii) (x-2)(x+2)(x-3) " " (iii) (x+2)(x+3)(x-5)` `(iv) (x-2)(x+3)(x-1) " " (v) (x-1)(x+1)(x-2)(x+2) " " (vi) (x+3)(2x-1)(x-2)`

Discussion

180.	If one zero of the polynomial p(x) = x3 – 6x2 + 11x – 6 is 3, find the other two zeroes.
Answer» p(x) = x³ – 6x² + 11x – 6 and its factor, x + 3 Let us divide p(x) by (x – 3). Here, x³ – 6x² + 11x – 6 = (x – 3) (x² – 3x + 2) = (x – 3) [(x² – (2 + 1) x + 2] = (x – 3) (x² – 2x – x + 2) = (x – 3) [x (x – 2) – 1(x – 2)] = (x – 3) (x – 1) (x – 2) ∴The other two zeroes are 1 and 2.

180.

If one zero of the polynomial p(x) = x3 – 6x2 + 11x – 6 is 3, find the other two zeroes.

Answer»

p(x) = x³ – 6x² + 11x – 6 and its factor, x + 3

Let us divide p(x) by (x – 3).

Here, x³ – 6x² + 11x – 6 = (x – 3) (x² – 3x + 2)

= (x – 3) [(x² – (2 + 1) x + 2]

= (x – 3) (x² – 2x – x + 2)

= (x – 3) [x (x – 2) – 1(x – 2)]

= (x – 3) (x – 1) (x – 2)

∴The other two zeroes are 1 and 2.

Discussion

181.	Show that x = 2 is a root of 2x3 + x2 - 7x - 6.
Answer» p(x) = 2x³ + x² - 7x - 6 then, p(2) = 2(2)³ + (2)² 7(2) - 6 = 16 + 4 - 14 - 6 = 0 Hence x = 2 is a root of p(x).

181.

Show that x = 2 is a root of 2x3 + x2 - 7x - 6.

Answer»

p(x) = 2x³ + x² - 7x - 6 then,

p(2) = 2(2)³ + (2)² 7(2) - 6 = 16 + 4 - 14 - 6 = 0

Hence x = 2 is a root of p(x).

Discussion

182.	Expand using suitable identities : (x + 2y + 4z)2
Answer» (x + 2y + 4z)² = (x)² + (2y)² + (4z)² + 2(x) (2y) + 2 (2y) (4z) + 2 (4z) (x) [ ∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx] = x² + 4y² + 16z² + 4xy + 16yz + 8zx

182.

Expand using suitable identities : (x + 2y + 4z)2

Answer»

(x + 2y + 4z)²

= (x)² + (2y)² + (4z)² + 2(x) (2y) + 2 (2y) (4z) + 2 (4z) (x)

[ ∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]

= x² + 4y² + 16z² + 4xy + 16yz + 8zx

Discussion

183.	The coefficient of x3 in x2 + 5 x – 1 is A) 5 B) – 1 C) 0 D) 72
Answer» Correct option is C) 0 Correct option is (C) 0 Given polynomial is \(x^2+5x-1\) \(\because\) There is no term with \(x^3\) as variable. \(\therefore\) Coefficient of \(x^3\) in \(x^2+5x-1\) is 0.

183.

The coefficient of x3 in x2 + 5 x – 1 is A) 5 B) – 1 C) 0 D) 72

Answer»

Correct option is C) 0

Correct option is (C) 0

Given polynomial is \(x^2+5x-1\)

\(\because\) There is no term with \(x^3\) as variable.

\(\therefore\) Coefficient of \(x^3\) in \(x^2+5x-1\) is 0.

Discussion

184.	The coefficient of x2 in 7x3 – 2x2 + 3x + 1 is A) 7 B) – 2 C) 3 D) 1
Answer» Correct option is (B) –2 The coefficient of \(x^2\) in \(7x^3-2x^2+3x+1\) is -2. Correct option is B) – 2

184.

The coefficient of x2 in 7x3 – 2x2 + 3x + 1 is A) 7 B) – 2 C) 3 D) 1

Answer»

Correct option is (B) –2

The coefficient of \(x^2\) in \(7x^3-2x^2+3x+1\) is -2.

Correct option is B) – 2

Discussion

185.	If p(x) = 2x2 – 5√3 x + 5, then find the value of p(5√3 ).
Answer» p(x) = 2x² – 5√3 x + 5 ∴ p(5√3) = 2(5√3)² – 5√3 (5√3 ) + 5 = 2 (25 x 3) – 25 x 3 + 5 = 150-75 + 5 ∴ p( 5√3 ) = 80

185.

If p(x) = 2x2 – 5√3 x + 5, then find the value of p(5√3 ).

Answer»

p(x) = 2x² – 5√3 x + 5

∴ p(5√3) = 2(5√3)² – 5√3 (5√3 ) + 5

= 2 (25 x 3) – 25 x 3 + 5

= 150-75 + 5

∴ p( 5√3 ) = 80

Discussion

186.	Verify whether the values of x is zero of the polynomial or not ?p(x) = (x – 1) (x + 2); x = – 1, – 2
Answer» The value of p(x) at x = – 1 is p(-1) = (-1 – 1) (-1 + 2) =-2 x 1 =-2 ≠ 0 Hence x = – 1 is not a zero of p(x). And the value of p(x) at x = – 2 is p (- 2) = (- 2 – 1) (- 2 + 2) = – 3 x 0 = 0 Hence, x = – 2 is a zero of p(x).

186.

Verify whether the values of x is zero of the polynomial or not ?p(x) = (x – 1) (x + 2); x = – 1, – 2

Answer»

The value of p(x) at x = – 1 is

p(-1) = (-1 – 1) (-1 + 2)

=-2 x 1 =-2 ≠ 0

Hence x = – 1 is not a zero of p(x).

And the value of p(x) at x = – 2 is

p (- 2) = (- 2 – 1) (- 2 + 2) = – 3 x 0 = 0

Hence, x = – 2 is a zero of p(x).

Discussion

187.	Verify whether the values of x is zero of the polynomial or not ?p(x) = x2 – 1; x = ±1
Answer» The value of p(x) at x = 1 and – 1 is p(1) = 1² – 1 = 0 p(-1) = (-1)² -1 = 0 ∴ x = ±1 is a zero of p(x).

187.

Verify whether the values of x is zero of the polynomial or not ?p(x) = x2 – 1; x = ±1

Answer»

The value of p(x) at x = 1 and – 1 is

p(1) = 1² – 1 = 0

p(-1) = (-1)² -1 = 0

∴ x = ±1 is a zero of p(x).

Discussion

188.	`2y^3+y^2-2y-1`
Answer» Here, `p(y) = 2y^3+y^2-2y-1` `=y^2(2y+1)-1(2y+1)` `= (y^2-1)(2y+1)` `=(y+1)(y-1)(2y+1)` So, zeroes of `p(y)` will be `1,-1 and -1/2.`

Discussion

189.	If 2 is a zero of the polynomial p(x) = 2x2 – 3x + 7a, find the value of a.
Answer» Given that 2 is a zero of p(x) = 2x² – 3x + 7a (i.e.) p(2) = 0 ⇒ 2(2)² – 3(2) + 7a = 0 ⇒ 8 – 6 + 7a = 0 ⇒ 2 + 7a = 0 ⇒ 7a = – 2 ⇒ a = -2/7

189.

If 2 is a zero of the polynomial p(x) = 2x2 – 3x + 7a, find the value of a.

Answer»

Given that 2 is a zero of p(x) = 2x² – 3x + 7a

(i.e.) p(2) = 0

⇒ 2(2)² – 3(2) + 7a = 0

⇒ 8 – 6 + 7a = 0

⇒ 2 + 7a = 0

⇒ 7a = – 2

⇒ a = -2/7

Discussion

190.	Find the remainder when x3 + 3x2 + 3x + 1 is divided by x + 1 Each Linear polynomials
Answer» Let f(x) = x³ + 3x² + 3x + 1 By remainder theorem, the remainder is f (- 1) f (- 1) = (- 1)³ + 3(- 1)² + 3(- 1) + 1 = – 1 + 3 – 3 + 1 = 0

190.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by x + 1 Each Linear polynomials

Answer»

Let f(x) = x³ + 3x² + 3x + 1

By remainder theorem, the remainder is f (- 1)

f (- 1) = (- 1)³ + 3(- 1)² + 3(- 1) + 1

= – 1 + 3 – 3 + 1 = 0

Discussion

191.	Expand each of the following, using suitable identities:(i) `(x+2y+4z)^2` (ii) `(2x-y+z)^2` (iii) `(-2x+3y+2z)^2` (iv) `(3a-7b-c)^2`(v) `(-2x+5y-3z)^2` (vi) `[1/4a-1/2b+1]^2`
Answer» Here, we will use the identity, `(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca` (i) `(x+2y+4z)^2 = x^2+(2y)^2+(4z)^2+2(x)(2y)+2(2y)(4z)+2(4z)(x)` `=x^2+4y^2+16z^2+4xy+16yz+8zx` (ii) `(2x-y+z)^2 = (2x)^2+(-y)^2+(z)^2+2(2x)(-y)+2(-y)(z)+2(z)(2x)` `=4x^2+y^2+z^2-4xy-2yz+4zx` Similarly, we can solve the other parts.

Discussion

192.	What is the coefficient form of x3 – 1 ? (A) (1, -1) (B) (3, -1) (C) (1, 0, 0, -1) (D) (1, 3, -1)
Answer» Correct option is (C) (1, 0, 0, -1) sol: x³ – 1 = x³ + 0x² + 0x – 1

192.

What is the coefficient form of x3 – 1 ? (A) (1, -1) (B) (3, -1) (C) (1, 0, 0, -1) (D) (1, 3, -1)

Answer»

Correct option is (C) (1, 0, 0, -1)

sol: x³ – 1 = x³ + 0x² + 0x – 1

Discussion

193.	If zetoes ofthe polynomia x2 + 4x + 2a are α and 2/α, then find the value of a.
Answer» Product of (zeroes) roots = c/a = 2a/1 = α. 2/α or 2a = 2 a = 1

193.

If zetoes ofthe polynomia x2 + 4x + 2a are α and 2/α, then find the value of a.

Answer»

Product of (zeroes) roots = c/a

= 2a/1 = α. 2/α

or

2a = 2

a = 1

Discussion

194.	Prove that if `p=2-a` , then `a^3+6ap+p^3-8=0`
Answer» P=a `a^3+6ap+p^3-8=0` LHS `a^3+6ap+p^3-8` `a^3+6a(2-a)+(2-a)^3-8` `a^3+6a*2-a)+8-a^3-6a(2-a)-8` ``0` RHS.

Discussion

195.	If x = 2 & x = 0 are two roots of the polynomial f(x) = 2x3 - 5x2 + ax + b. Find the values of and b.
Answer» f(x) = 2(2)³ - 5(2)² + a(2) + b = 0 ⇒ 16 - 20 + 2a + b = 0 ⇒ 2a + b = 4 ....(i) ⇒ f(0) = 2(0)³ - 5(0)² + a(0) + b = 0 ⇒ b = 0 So, 2a = 4 Hence, a = 2, b = 0

195.

If x = 2 & x = 0 are two roots of the polynomial f(x) = 2x3 - 5x2 + ax + b. Find the values of and b.

Answer»

f(x) = 2(2)³ - 5(2)² + a(2) + b = 0

⇒ 16 - 20 + 2a + b = 0

⇒ 2a + b = 4 ....(i)

⇒ f(0) = 2(0)³ - 5(0)² + a(0) + b = 0

⇒ b = 0

So, 2a = 4

Hence, a = 2, b = 0

Discussion

196.	Verify whether the values of x is zero of the polynomial or not ?p(x) = ax + b ; x = -b/a
Answer» The value of p(x) at x = -b/a is \(p (\frac {-b}{a}) = a (\frac {-b}{a}) + b\) = -b + b = 0 ∴ x = -b/a is a zero of p (x)

196.

Verify whether the values of x is zero of the polynomial or not ?p(x) = ax + b ; x = -b/a

Answer»

The value of p(x) at x = -b/a is

\(p (\frac {-b}{a}) = a (\frac {-b}{a}) + b\)

= -b + b = 0

∴ x = -b/a is a zero of p (x)

Discussion

197.	If 0 and 1 are the zeroes of the polynomial f(x) = 2x3 – 3x2 + ax + b, find the values of a and b.
Answer» Given that f(0) = 0; f(1) = 0 and f(x) = 2x³ – 3x² + ax + b ∴ f(0) = 2(0)³ – 3(0)² + a(0) + b ⇒ 0 = b Also f(1) = 0 ⇒ 2(1)³ – 3(1)² + a(1) + 0 = 0 ⇒ 2 – 3 + a = 0. ⇒ a = 1 Hence a = 1; b = 0

197.

If 0 and 1 are the zeroes of the polynomial f(x) = 2x3 – 3x2 + ax + b, find the values of a and b.

Answer»

Given that f(0) = 0; f(1) = 0 and

f(x) = 2x³ – 3x² + ax + b

∴ f(0) = 2(0)³ – 3(0)² + a(0) + b

⇒ 0 = b

Also f(1) = 0

⇒ 2(1)³ – 3(1)² + a(1) + 0 = 0

⇒ 2 – 3 + a = 0.

⇒ a = 1

Hence a = 1; b = 0

Discussion

198.	If `x+y+z=0` show that `x^3+y^3+z^3=3x y z`.
Answer» We have this formula, `x^3 + y^3 +z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)` Here, we are given, `x+y+z = 0`, So,our formula becomes `x^3 + y^3 +z^3 - 3xyz =0**(x^2+y^2+z^2-xy-yz-zx)` `x^3 + y^3 +z^3 - 3xyz =0` `x^3 + y^3 +z^3 = 3xyz `

Discussion

199.	Find the value of p(x) = 3x2 + x + 7 when x = – 2.
Answer» p(x) = 3x² + x + 7 Substituting x = – 2, we get p(-2) = 3(2)² + (-2) + 7 = 12 – 2 + 7 ∴ p(-2) = 17

199.

Find the value of p(x) = 3x2 + x + 7 when x = – 2.

Answer»

p(x) = 3x² + x + 7

Substituting x = – 2, we get

p(-2) = 3(2)² + (-2) + 7

= 12 – 2 + 7

∴ p(-2) = 17

Discussion

200.	What is the degree of the polynomial ?(A) 0 (B) 1 (C) undefined (D) any real number
Answer» Correct option is (C) undefined

Discussion

Explore topic-wise InterviewSolutions in .

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Which of the following expressions is a polynomial ?A) 3x-2 + 7x + 1B) 7x2 + 2/x – 8C) x2 – x – 1D) x +x/2 + 1/x

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Which of the following is not a zero of the polynomial x3 – 6x2 + 11x – 6? A) 1 B) 2 C) 3 D) 0

Using factor theorem; factorize the polynomial `x^3-6x^2+11x-6`

If one zero of the polynomial p(x) = x3 – 6x2 + 11x – 6 is 3, find the other two zeroes.

Show that x = 2 is a root of 2x3 + x2 - 7x - 6.

Expand using suitable identities : (x + 2y + 4z)2

The coefficient of x3 in x2 + 5 x – 1 is A) 5 B) – 1 C) 0 D) 72

The coefficient of x2 in 7x3 – 2x2 + 3x + 1 is A) 7 B) – 2 C) 3 D) 1

If p(x) = 2x2 – 5√3 x + 5, then find the value of p(5√3 ).

Verify whether the values of x is zero of the polynomial or not ?p(x) = (x – 1) (x + 2); x = – 1, – 2

Verify whether the values of x is zero of the polynomial or not ?p(x) = x2 – 1; x = ±1

`2y^3+y^2-2y-1`

If 2 is a zero of the polynomial p(x) = 2x2 – 3x + 7a, find the value of a.

Find the remainder when x3 + 3x2 + 3x + 1 is divided by x + 1 Each Linear polynomials

Expand each of the following, using suitable identities:(i) `(x+2y+4z)^2` (ii) `(2x-y+z)^2` (iii) `(-2x+3y+2z)^2` (iv) `(3a-7b-c)^2`(v) `(-2x+5y-3z)^2` (vi) `[1/4a-1/2b+1]^2`

What is the coefficient form of x3 – 1 ? (A) (1, -1) (B) (3, -1) (C) (1, 0, 0, -1) (D) (1, 3, -1)

If zetoes ofthe polynomia x2 + 4x + 2a are α and 2/α, then find the value of a.

Prove that if `p=2-a` , then `a^3+6ap+p^3-8=0`

If x = 2 &amp; x = 0 are two roots of the polynomial f(x) = 2x3 - 5x2 + ax + b. Find the values of and b.

Verify whether the values of x is zero of the polynomial or not ?p(x) = ax + b ; x = -b/a

If 0 and 1 are the zeroes of the polynomial f(x) = 2x3 – 3x2 + ax + b, find the values of a and b.

If `x+y+z=0` show that `x^3+y^3+z^3=3x y z`.

Find the value of p(x) = 3x2 + x + 7 when x = – 2.

What is the degree of the polynomial ?(A) 0 (B) 1 (C) undefined (D) any real number

If x = 2 & x = 0 are two roots of the polynomial f(x) = 2x3 - 5x2 + ax + b. Find the values of and b.