Factor and Remainder Theorems

$\left(x - 1\right)$ is a factor of the expression $2x^3 + ax^2 + bx - 14$.

When the expression is divided by $\left(x - 3\right)$, it leaves a remainder $52$.

Find the values of $a$ and $b$.

Let $\;$ $f \left(x\right) = 2x^3 + ax^2 + bx -14$

Given: $\;$ $\left(x - 1\right)$ is a factor of $f \left(x\right)$

Then, by factor theorem, $\;$ $f \left(1\right) = 0$

i.e. $\;$ $f \left(1\right) = 2 \times 1^3 + a \times 1^2 + b \times 1 - 14 = 0$

i.e. $\;$ $a + b = 12$ $\;\;\; \cdots \; (1)$

Given: $\;$ When $f \left(x\right)$ is divided by $\left(x - 3\right)$, it leaves a remainder $52$

Then, by remainder theorem, $\;$ $f \left(3\right) = 52$

i.e. $\;$ $f \left(3\right) = 2 \times 3^3 + a \times 3^2 + b \times 3 - 14 = 52$

i.e. $\;$ $9a + 3b = 12$

i.e. $\;$ $3a + b = 4$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously, we have

$2a = -8$ $\implies$ $a = -4$

Substituting the value of $a$ in equation $(1)$ gives

$b = 12 - a = 12 - \left(-4\right) = 16$

Factor and Remainder Theorems

Use the remainder theorem to factorize the expression $\;$ $2x^3 + x^2 - 13x + 6$

Let $\;$ $f \left(x\right) = 2x^3 + x^2 - 13x + 6$

For $x = 2$, the value of the given expression is:

$f \left(2\right) = 2 \times 2^3 + 2^2 - 13 \times 2 + 6 = 16 + 4 - 26 + 6 = 0$

$\implies$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$.

$\begin{array}{lllll} x - 2 & ) & 2x^3 + x^2 - 13x + 6 & ( & 2x^2 + 5x - 3 \\ & & 2x^3 - 4x^2 & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ 5x^2 - 13x & & \\ & & \ \ \ \ \ \ \ \ 5x^2 - 10x & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3x + 6 & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3x + 6 & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \end{array}$

$\begin{aligned} \therefore \; 2x^3 + x^2 - 13x + 6 & = \left(x - 2\right) \left[2x^2 + 5x - 3\right] \\\\ & = \left(x - 2\right) \left[2x^2 + 6x - x - 3\right] \\\\ & = \left(x - 2\right) \left[2x \left(x + 3\right) - 1 \left(x + 3\right)\right] \\\\ & = \left(x - 2\right) \left(2x - 1\right) \left(x + 3\right) \end{aligned}$

Factor and Remainder Theorems

If $x^3 + ax^2 -x + b$ has $\left(x - 2\right)$ as a factor and leaves a remainder $3$ when divided by $\left(x - 3\right)$, find $a$ and $b$.

Let $f \left(x\right) = x^3 + ax^2 -x + b$

$\because \;$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$,

$\therefore \;$ by factor theorem $f \left(2\right) = 0$

Now, $\;$ $f \left(2\right) = 2^3 + a \times 2^2 - 2 + b = 8 + 4a -2 + b = 6 + 4a + b$

$\therefore \;$ $f \left(2\right) = 0$ $\implies$ $6 + 4a + b = 0$

i.e. $\;$ $4a + b = -6$ $\;\;\; \cdots \; (1)$

$\because \;$ $f \left(x\right)$ when divided by $\left(x - 3\right)$ leaves a remainder of $3$,

$\therefore \;$ by remainder theorem $f \left(3\right) = 3$

Now, $\;$ $f \left(3\right) = 3^3 + a \times 3^2 -3 + b = 27 + 9a -3 + b = 24 + 9a + b$

$\because \;$ $f \left(3\right) = 3$ $\implies$ $24 + 9a + b = 3$

i.e. $\;$ $9a + b = -21$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously gives

$5a = -15$ $\implies$ $a = -3$

and $\;$ $b = -6 - 4a = -6 -4 \times \left(-3\right) = 6$

$\therefore \;$ $a = -3, \; b = 6$

Factor and Remainder Theorems

When the two polynomials $x^3 + ax^2 -x -2$ and $ax^3 + x^2 -6x -4$ are divided by $x - 2$, the remainder is the same. Find the value of $a$.

Let $\;$ $f \left(x\right) = x^3 + ax^2 - x - 2$ $\;$ and $\;$ $g \left(x\right) = ax^3 + x^2 -6x -4$

By remainder theorem,

when $f \left(x\right)$ is divided by $\left(x - 2\right)$, the remainder is $= f \left(2\right)$

when $g \left(x\right)$ is divided by $\left(x - 2\right)$, the remainder is $= g \left(2\right)$

Given: $\;$ $f \left(2\right) = g \left(2\right)$ $\;\;\; \cdots \; (1)$

Now, $\;$ $f \left(2\right) = 2^3 + a \times 2^2 - 2 - 2 = 4 + 4a$ $\;\;\; \cdots \; (2)$

$g \left(2\right) = a \times 2^3 + 2^2 - \left(6 \times 2\right) - 4 = 8a - 12$ $\;\;\; \cdots \; (3)$

$\therefore \;$ In view of equations $(2)$ and $(3)$ equation $(1)$ becomes,

$4 + 4a = 8a - 12$

i.e. $\;$ $4a = 16$ $\implies$ $a = 4$

Factor and Remainder Theorems

The polynomials $2x^3 - 7x^2 + ax - 6$ and $x^3 - 8x^2 + \left(2a + 1 \right)x - 16$ leave the same remainder when divided by $\left(x - 2\right)$. Find the value of $a$.

Let $\;$ $f \left(x\right) = 2x^3 - 7x^2 + ax - 6$, $\;$ $g \left(x\right) = x^3 - 8x^2 + \left(2a + 1\right)x - 16$

By remainder theorem,

remainder obtained when $f \left(x\right)$ is divided by $\left(x - 2\right)$ is $ = f \left(2\right)$

remainder obtained when $g \left(x\right)$ is divided by $\left(x - 2\right)$ is $ = g \left(2\right)$

Given: $\;$ $f \left(2\right) = g \left(2\right)$ $\;\;\; \cdots \; (1)$

Now, $\;$ $f \left(2\right) = 2 \times 2^3 - 7 \times 2^2 + 2a - 6$

i.e. $\;$ $f \left(2\right) = 2a - 18$ $\;\;\; \cdots \; (2a)$

and $\;$ $g \left(2\right) = 2^3 - 8 \times 2^2 + \left(2a + 1\right) \times 2 - 16$

i.e. $\;$ $g \left(x\right) = 4a - 38$ $\;\;\; \cdots \; (2b)$

In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes,

$2a - 18 = 4a - 38$

i.e. $\;$ $2a = 20$ $\implies$ $a = 10$

Factor and Remainder Theorems

If $\;$ $2x^3 + ax^2 + bx - 2$ $\;$ when divided by $\;$ $2x - 3$ $\;$ and $\;$ $x + 3$ $\;$ leaves remainder $7$ and $-20$ respectively, find the values of $a$ and $b$.

Let $f \left(x\right) = 2x^3 + ax^2 + bx - 2$

Given: $\;$ $f \left(x\right)$ when divided by $2x - 3$ leaves remainder $7$

$\therefore \;$ By remainder theorem, $\;$ $f \left(\dfrac{3}{2}\right) = 7$

i.e. $\;$ $2 \times \left(\dfrac{3}{2}\right)^3 + a \times \left(\dfrac{3}{2}\right)^2 + b \times \dfrac{3}{2} - 2 = 7$

i.e. $\;$ $\dfrac{27}{4} + \dfrac{9a}{4} + \dfrac{3b}{2} = 9$

i.e. $\;$ $9a + 6b = 36 - 27 = 9$

i.e. $\;$ $3a + 2b = 3$ $\;\;\; \cdots \; (1)$

Given: $\;$ $f \left(x\right)$ when divided by $x + 3$ leaves remainder $-20$

$\therefore \;$ By remainder theorem, $\;$ $f \left(-3\right) = -20$

i.e. $\;$ $2 \times \left(-3\right)^3 + a \times \left(-3\right)^2 + b \times \left(-3\right) - 2 = -20$

i.e. $\;$ $9a - 3b = -18 + 54 = 36$

i.e. $\;$ $3a - b = 12$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously, we get,

$3b = -9$ $\implies$ $b = -3$

Substituting the value of $b$ in equation $(2)$, we get,

$3a + 3 = 12$

i.e. $\;$ $3a = 9$ $\implies$ $a = 3$

$\therefore \;$ $a = 3$, $\;$ $b = -3$

Factor and Remainder Theorems

The expression $x^3 + ax^2 + bx + 6$ has $\left(x - 2\right)$ as a factor and leaves a remainder $3$ when divided by $\left(x - 3\right)$. Find '$a$' and '$b$'.

Let $\;$ $f \left(x\right) = x^3 + ax^2 + bx + 6$

Given: $\;$ $\left(x - 2\right)$ is a factor of $f\left(x\right)$

Then, by factor theorem, $\;$ $f \left(2\right) = 0$

i.e. $\;$ $2^3 + a \times 2^2 + b \times 2 + 6 = 0$

i.e. $\;$ $4a + 2b + 14 = 0$

i.e. $\;$ $2a + b = - 7$ $\;\;\; \cdots \; (1)$

Given: $\;$ $f \left(x\right)$ when divided by $\left(x - 3\right)$ leaves a remainder $3$.

Then, by remainder theorem, $\;$ $f \left(3\right) = 3$

i.e. $\;$ $3^3 + a \times 3^2 + b \times 3 + 6 = 3$

i.e. $\;$ $9a + 3b + 12 = 0$

i.e. $\;$ $3a + b = - 4$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously we have,

$a = 3$ $\;$ and $\;$ $b = - 13$

From Exam Papers

Remainder Theorem

Using remainder theorem show that $\left(2x + 1\right)$ is a factor of the polynomial $p \left(x\right) = 4x^3 + 4x^2 - x - 1$. Hence factorize the polynomial.

Given: $p \left(x\right) = 4x^3 + 4 x^2 - x -1$

Now, $\;$ $2x + 1 =0$ $\implies$ $x = - \dfrac{1}{2}$

Remainder $=$ Value of $p \left(x\right)$ at $x = - \dfrac{1}{2}$

$\begin{aligned} p \left(- \dfrac{1}{2}\right) & = 4 \left(- \dfrac{1}{2}\right)^3 + 4 \left(- \dfrac{1}{2}\right)^2 - \left(- \dfrac{1}{2}\right) - 1 \\\\ & = - \dfrac{1}{2} + 1 + \dfrac{1}{2} - 1 \\\\ & = 0 \end{aligned}$

$\implies$ $\left(2x + 1\right)$ is a factor of $p \left(x\right)$

Now,

$\begin{array}{lllll} 2x + 1 & ) & 4x^3 + 4x^2 - x - 1 & ( & 2x^2 + x - 1 \\ & & 4x^3 + 2x^2 & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ 2x^2 - x & & \\ & & \ \ \ \ \ \ \ \ 2x^2 + x & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2x - 1 & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2x - 1 & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \end{array}$

$\begin{aligned} \therefore \; 4 x^3 + 4 x^2 - x - 1 & = \left(2x + 1\right) \left(2x^2 + x -1\right) \\\\ & = \left(2x + 1\right) \left(2x^2 + 2 x - x - 1\right) \\\\ & = \left(2x + 1\right) \left[2 x \left(x + 1\right) - 1 \left(x + 1\right)\right] \\\\ & = \left(2x + 1\right) \left(2x - 1\right) \left(x + 1\right) \end{aligned}$