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 Theorem

Find the value of $a$ and $b$ if $\left(x - 1\right)$ and $\left(x - 2\right)$ are factors of $x^3 - ax + b$.

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

$\because \;$ $\left(x - 1\right)$ is a factor of $f \left(x\right)$ $\implies$ by factor theorem $\;$ $f \left(1\right) = 0$

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

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

$\because \;$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$ $\implies$ by factor theorem $\;$ $f \left(2\right) = 0$

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

i.e. $\;$ $2a - b = 8$ $\;\;\; \cdots \; (2)$

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

$a = 7$ $\;$ and $\;$ $b = 6$

Factor Theorem

Use the factor theorem to factorize completely: $\;$ $x^3 + x^2 -4x - 4$

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

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

$\implies$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$ $\;\;\;$ [By factor theorem]

Now,

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

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

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 $\;$ $b = - 13$