Class 10: Remainder and Factor Theorems – Misc. Problems – Set 1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the remainder when $x^2 - 8x + 4$ is divided by $2x + 1.$

Answer:

$\displaystyle \text{Step 1} : 2x + 1 = 0 \Rightarrow x = \frac{-1}{2}$

$\displaystyle \text{Step 2: Required remainder = Value of given polynomial } x^2 - 8x + 4 \\ \\ \text{ at } x = \frac{-1}{2}$

$\displaystyle \text{Therefore Remainder } = \Big(\frac{-1}{2} \Big)^2 - 8 \Big( \frac{-1}{2} \Big) + 4 = \frac{1}{4}+4+4=8\frac{1}{4}$

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Question 2: Find the value of $'a'$ if the division of $ax^3 + 9x^2 + 4x - 10$ by $x + 3$ leaves a remainder of 5.

Answer:

$x+3=0 \Rightarrow x=-3$

$\text{Given, remainder is 5; therefore: }$

The value of $ax^3 + 9x^2 + 4x - 10$ at $x =-3$ is $5$

$\Rightarrow a(-3)^3 + 9(-3)^2 + 4(-3) - 10 = 5$

$\Rightarrow -27a+ 81-12-10=5 \text{ or } a=2$

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Question 3: When the polynomial $2x^3 -kx^2 + (5k- 3)x- 8$ is divided by, $x- 2,$ the remainder is 14. Find the value of $'k'.$

Answer:

$x-2 = 0 \Rightarrow x = 2$

Given. remainder is 14, therefore:

$\Rightarrow 2(2)^3 - k(2)^2 + ( 5k-3) \times 2 - 8 = 14$

$\Rightarrow 16 - 4k +10k - 6-8 = 14$

$\Rightarrow 6k = 12$

$\Rightarrow k = 2$

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Question 4: The polynomials $3x^3 - ax^2 + 5x- 13$ and $(a + 1)x^2 -7x+ 5$ leave the same remainder when divided by $x - 3.$ Find the value of $'a'.$

Answer:

$x-3 =0 \Rightarrow x=3$

Since, the given polynomials leave the same remainder when divided by $x - 3$

Value of polynomial $3x^3 - 9x^2 + 5x - 13 \text{ at } x = 3$ is the same as the value of polynomial $(a + 1) x^2 - 7x + 5 \text{ at } x = 3$

$\Rightarrow 3(3)^3 - 9(3)^2 + 5 \times 3 - 13 = ( a+1)(3)^2 - 7 \times 3 + 5$

$\Rightarrow 81 - 9a + 15 - 13 = 9a + 9 - 21 + 5$

$\Rightarrow 18a = 90$

$\Rightarrow a = 5$

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Question 5: When $f(x) = x^3 + ax^2 - bx - 8$ is divided by $x - 2,$ the remainder is zero and when divided by $x + 1,$ the remainder is $-30.$ Find the values of $a$ and $b.$

Answer:

Since, $x - 2 = 0 \Rightarrow x = 2$

Given that on dividing $f(x) = x^3 + ax^2 - bx - 8 \text{ by } x - 2$ the remainder $= 0$

$\Rightarrow f(2) = 0$

$\Rightarrow (2)^3 + a(2)^2 - b(2) - 8 = 0$

$\Rightarrow 8 + 4a - 2b - 8 = 0$

$\Rightarrow 4a - 2b = 0$

$\Rightarrow 2a - b = 0 \text{ ... ... ... ... ... i)}$

Given that on dividing $f(x) = x^3 + ax^2 - bx - 8 \text{ by } x +1$ the remainder $= -30$

$\Rightarrow f(-1) = -30$

$\Rightarrow (-1)^3 + a( -1)^2 - b(-1) - 8 = -30$

$\Rightarrow a+b = 21 \text{ ... ... ... ... ... ii)}$

Solving i) and ii) we get $a = -7$ and $b = -14$

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Question 6: What number should be added to $2x^3 - 3x^2 + x$ so that when the resulting polynomial is divided by $x - 2,$ the remainder is $3$ ?

Answer:

Let the number added be $k$ so the resulting polynomial is $2x^3 - 3x^2 + x + k$

Given, when this polynomial is divided by $x- 2,$ the remainder $= 3$

$\Rightarrow 2(2)^3 - 3 ( 2)^2 + 2 + k = 3$

$\Rightarrow 16-12+2+k = 3$

$\Rightarrow k = - 3$

Therefore the required number to be added $= - 3$

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Question 7: Determine whether $x-1$ is a factor of $x^6 - x^5 + x^4 + x^3 - x^2 - x + 1$ or not?

Answer:

$x-1 = 0 \Rightarrow x = 1$

Therefore when given polynomial is divided by $x-1,$ the remainder

$= (1)^6 - (1)^5 + (1)^4 + (1)^3 - (1)^2 - (1) + 1 = 1 - 1 + 1 +1-1-1+1 = 4-3 = 1$

which is not equal to zero.

Therefore $x-1$ is not a factor of the given polynomial.

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Question 8: If $x- 2$ is a factor of $x^2 - 7x + 2a,$ find the value of a.

Answer:

$x-2 = 0 \Rightarrow x = 2$

Since, $x-2$ is a factor of polynomial $x^2 - 7x + 2a$

Therefore remainder $= 0$

$\Rightarrow (2)^2 - 7(2) + 2a =0$

$\Rightarrow a = 5$

$\\$

Question 9: Find the value of $'k'$ if $(x -2)$ is a factor of $x^3 + 2x^2 - kx +10.$ Hence, determine whether $(x + 5)$ is also a factor. [ICSE Board 2011]

Answer:

$x - 2$ is a factor and $x - 2 = 0 \Rightarrow x = 2$

Therefore The value of given expression $x^3 + 2x^2 - kx +10$ is zero at $x = 2$

$\Rightarrow \text{ Remainder } = 0$

$\Rightarrow (2)^3 + 2(2)^2 - k \times 2 + 10 = 0$

$\Rightarrow 8 + 8 -2k + 10 = 0$

$\Rightarrow k = 13$

On substituting $k = 13,$ the given expression becomes $x^3 + 2x^2 - 13x +10.$

Now to check whether $(x + 5)$ is also a factor or not, find the value of the given expression for $x =-5$

Therefore $x^3 + 2x^2 - 13x +10 \text{ at } x = -5$

$= (-5)^3 + 2(-5)^2 - 13(-5) +10 = -125 + 50 + 65 + 10 = -125 + 125 =0$

Since, the remainder is $0$

$\Rightarrow (x+ 5)$ is a factor.

$\\$

Question 10: Given that $x + 2$ and $x - 3$ are factors of $x^3 + ax + b;$ calculate the values of $a$ and $b.$

Answer:

Given, $x+ 2$ is a factor of $x^3 + ax + b;$

$\Rightarrow (-2)^3 + a(-2) + b = 0$

$\Rightarrow -2a + b = 8 \text{ ... ... ... ... ... i)}$

Again, given that : $x-3$ is a factor of $x^3 + ax + b;$

$\Rightarrow (3)^3 + a(3) + b = 0$

$\Rightarrow 3a+b = -27 \text{ ... ... ... ... ... ii)}$

Solving i) and ii) we get $a = -7$ and $b = -6$

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Question 11: Polynomial $x^3 - ax ^2 + bx - 6$ leaves remainder $-8$ when divided by $x-1$ and $x-2$ is a factor of it. Find the values of $a$ and $b.$

Answer:

On dividing by $x -1 ,$ the polynomial $x^3 - bx^2 + bx - 6$ leaves remainder $-8.$

$\Rightarrow (1)^3 - a(1)^2 + b(1) - 6 = - 8$

$\Rightarrow - a + b = - 3$

$\Rightarrow a - b = 3 \text{ ... ... ... ... ... i)}$

$(x-2)$ is a factor of polynomial $x^3 - bx^2 + bx - 6$

$\Rightarrow (2)^2 - a(2)^2 + b( 2) - 6 = 0$

$\Rightarrow 8 - 4a + 2b - 6 = 0$

$\Rightarrow 2a - b = 1 \text{ ... ... ... ... ... ii)}$

Solving i) and ii) we get $a = -2$ and $b = -5$

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Question 12: Using the Factor Theorem, show that $(x - 2)$ is a factor of $3x^2-5x-2.$ Hence, factorize the given expression.

Answer:

Since $x-2=0$

$\Rightarrow x=2$

Therefore Remainder = The value of $3x^2-5x-2$ at $x = 2$

$\begin{array}{rll} x-2& \overline{)3x^2-5x-2(} &3x+1 \\ (-) & \underline{3x^2-6x} & \\ & {\hspace{1.0cm} x-2 } & \\ (-) & {\hspace{1.0cm} \underline{x-2 } } & \\ & {\hspace{1.5cm}\times } & \end{array}$

$= 3(2)^2 - 5(2) - 2 = 12 - 10 -2 = 0$

$\Rightarrow (x-2)$ is a factor of $3x^2 - 5x - 2$

Now dividing $( 3x^2 - 5x - 2)$ by $( x-2),$ we get quotient $= 3x + 1$

Therefore $3x^2 - 5x - 2 = (x-2)(3x+1)$

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Question 13: Show that $2x+7$ is a factor of $2x^3 + 5x^2 - 11x - 14.$ Hence factorize the given expression completely, using the factor theorem. [ICSE Board 2006].

Answer:

$\displaystyle 2x+7 = 0 \Rightarrow x = \frac{-7}{2}$

$\displaystyle \text{Remainder }= \text{ Value of } 2x^3 + 5x^2 - 11x - 14 \text{ at } x = \frac{-7}{2}$

$\displaystyle = 2\Big(\frac{-7}{2} \Big)^3 + 5\Big(\frac{-7}{2} \Big)^2 - 11\Big(\frac{-7}{2} \Big) - 14$

$\displaystyle = - \frac{343}{4} + \frac{245}{4} + \frac{77}{2} - 14$

$\displaystyle = \frac{-343+245+154-56}{4} = 0$

$\displaystyle \Rightarrow (2x+7) \text{ is a factor } \text{ of } 2x^3 + 5x^2 - 11x - 14$

$\begin{array}{rll} 2x+7& \overline{)2x^3 + 5x^2 - 11x - 14(} & x^2-x-2 \\ (-) & \underline{2x^3+7x^2} & \\ & {\hspace{1.0cm} -2x^2-11x-14 } & \\ (-) & {\hspace{1.0cm} \underline{-2x^2-7x } } & \\ & {\hspace{2.0cm}-4x-14 } & \\ (-) & {\hspace{2.0cm}\underline{-4x-14 } }& \\ & {\hspace{2.5cm}\times } & \end{array}$

$\displaystyle \therefore 2x^3 + 5x^2 - 11x - 14 = (2x+7)( x^2-x-2)$

$\displaystyle = ( 2x+7) ( x^2 - 2x + x - 2)$

$\displaystyle = ( 2x+7) [x(x-2)+ 1(x-2)]$

$\displaystyle = ( 2x+ 7) ( x-2)(x+1)$

$\\$

Question 14: Using the Remainder Theorem, factorize the expression $2x^3 + x^2 - 2x - 1$ completely.

Answer:

First Step : For $x = 1,$ the value of given expression

$= 2(1)^3 + (1)^2 - 2 ( 1) - 1 = 2 + 1 - 2 - 1 = 0$

$\Rightarrow x-1$ is a factor of $2x^3 + x^2 - 2x - 1$

$\begin{array}{rll} x-1& \overline{)2x^3 + x^2 - 2x - 1(} & 2x^2 + 3x + 1 \\ (-) & \underline{2x^3-2x^2} & \\ & {\hspace{1.0cm} 3x^2-3x-1 } & \\ (-) & {\hspace{1.0cm} \underline{3x^2-3x } } & \\ & {\hspace{2.0cm}x-1 } & \\ (-) & {\hspace{2.0cm}\underline{x-1 } }& \\ & {\hspace{2.5cm}\times } & \end{array}$

Second Step:

$2x^3 + x^2 - 2x - 1 = (x-1)( 2x^2 + 3x + 1)$

$= ( x-1) ( 2x^2 + 3x + 1)$

$= ( x-1)(2x^2 + 2x + x + 1)$

$= ( x-1)[ 2x( x+1)+1(x+1)]$

$= ( x-1)(x+1)(2x+1)$

$\\$

Question 15: Find the values of $a$ and $b$ so that the polynomial $x^3 + ax^2 + bx - 45$ has $(x - 1)$ and $(x + 5)$ as its factors. For the values of $a$ and $b$ , as obtained above, factorize the given polynomial completely.

Answer:

$(x - 1)$ is a factor of given polynomial $x^3 + ax^2 + bx - 45$

$\Rightarrow (1)^3 + a(1)^2 + b(1) - 45 = 0$

$\Rightarrow a+b = 44 \text{ ... ... ... ... ... i) }$

$(x +5)$ is a factor of given polynomial $x^3 + ax^2 + bx - 45$

$\Rightarrow (-5)^3 + a(-5)^2 + b(-5) - 45 = 0$

$\Rightarrow 5a- b = 34 \text{ ... ... ... ... ... ii) }$

On solving equations i) and ii), we get : $a = 13$ and $b = 31$

Therefore the given polynomial $x^3 + ax^2 + bx - 45 = x^3 + 13x^2 + 31x - 45$

Now divide this polynomial by $( x-1)$ we get

$\begin{array}{rll} x-1& \overline{)x^3+13x^2+31x-45(} & x^2+14x+45 \\ (-) & \underline{x^3-x^2} & \\ & {\hspace{1.0cm} 14x^2+31x-45 } & \\ (-) & {\hspace{1.0cm} \underline{14x^2-14x } } & \\ & {\hspace{2.0cm}45x-45 } & \\ (-) & {\hspace{2.0cm}\underline{45x-45 } }& \\ & {\hspace{2.5cm}\times } & \end{array}$

$x^3 + 13x^2 + 31x - 45 = (x-1)(x^2 + 14 x + 45)$

$= (x-1)( x^2 + 9x + 5x + 45)$

$= ( x-1)[ x(x+9) + 5 ( x + 9)]$

$= ( x-1)( x+9)(x+5)$

$\\$

Question 16: If $(x - 2)$ is a factor of $2x^3 - x^2 - px - 2$

(i) find the value of $p.$

(ii) with the value of $p,$ factorize the above expression completely

Answer:

(i) $x-2=0 \Rightarrow x=2$

Since, $(x - 2)$ is a factor of given expression $\Rightarrow \text{ Remainder } = 0$

$\Rightarrow 2(2)^3 - (2)^2 - p \times 2 - 2 = 0$

$\Rightarrow 10-2p = 0$

$\Rightarrow p = 5$

(ii) $\therefore 2x^3 - x^2 - px - 2 = 2x^3 - x^2 - 5x - 2$

On dividing $2x^3 - x^2 - 5x - 2$ by $(x-2)$ we get:

$\begin{array}{rll} x-2& \overline{)2x^3-x^2-5x-2(} & 2x^2 + 3x + 1 \\ (-) & \underline{2x^3-4x^2} & \\ & {\hspace{1.0cm} 3x^2-5x-2 } & \\ (-) & {\hspace{1.0cm} \underline{3x^2-6x } } & \\ & {\hspace{2.0cm}x-2 } & \\ (-) & {\hspace{2.0cm}\underline{x-2 } }& \\ & {\hspace{2.5cm}\times } & \end{array}$