Question

The polynomial $f (x)$ is defined by $f (x) = 12 x^{3} + 25 x^{2} - 4 x - 12$ .

Show that $f (- 2) = 0$ and factorise $f (x)$ completely.

Accepted Answer

Given $f (x) = 12 x^{3} + 25 x^{2} - 4 x - 12$

$\Rightarrow f (- 2) = 12 {(- 2)}^{3} + 25 {(- 2)}^{2} - 4 (- 2) - 12$

$\Rightarrow f (- 2) = 12 (- 8) + 25 (4) - 8 - 12$

$\Rightarrow f (- 2) = - 96 + 100 - 4$

$\Rightarrow f (- 2) = 0$

$(x + 2)$ , is a factor of $f (x)$ by the factor theorem.

Dividing the cubic by $x + 2$ , using long division.

Divide the first term of the polynomial by $x$ :

$12 x^{3} \div x = 12 x^{2}$ .

Multiply $(x + 2)$ by $12 x^{2}$ : $12 x^{2} (x + 2) = 12 x^{3} + 24 x^{2}$

Subtract: $(12 x^{3} + 25 x^{2}) - (12 x^{3} + 24 x^{2}) = x^{2}$

Bring down the $- 4 x$ , from the next column.

$\begin{matrix} 12 x^{2} \\ x - 2 12 x^{3} + 25 x^{2} - 4 x - 12 \\ \frac{12 x^{3} - 24 x^{2}}{x^{2} - 4 x} \end{matrix}$

Divide $x^{2}$ by $x$ : $x^{2} \div x = x$ .

Multiply $(x + 2)$ by $x$ : $x (x + 2) = x^{2} + 2 x$

Subtract: $(x^{2} - 4 x) - (x^{2} + 2 x) = - 6 x$

Bring down the $- 12$ , from the next column.

$\begin{matrix} 12 x^{2} + x \\ x + 2 12 x^{3} + 25 x^{2} - 4 x - 12 \\ \frac{12 x^{3} + 24 x^{2}}{x^{2} - 4 x} \\ \frac{x^{2} + 2 x}{- 6 x - 12} \end{matrix}$

Divide $- 6 x$ by $x$ : $- 6 x \div x = - 6$ .

Multiply $(x + 2)$ by $- 6$ : $(- 6) (x + 2) = - 6 x - 12$

Subtract: $(- 6 x - 12) - (- 6 x - 12) = 0$

$\begin{matrix} 12 x^{2} + x - 6 \\ x + 2 12 x^{3} + 25 x^{2} - 4 x - 12 \\ \frac{12 x^{3} + 24 x^{2}}{x^{2} - 4 x} \\ \frac{x^{2} + 2 x}{- 6 x - 12} \\ \frac{- 6 x - 12}{0} \end{matrix}$

Hence, $(12 x^{3} + 25 x^{2} - 4 x - 12) \div (x + 2) = 12 x^{2} + x - 6$

$\Rightarrow 12 x^{2} + x - 6 = 12 x^{2} - 8 x + 9 x - 6 = 0$ $\Rightarrow 4 x (3 x - 2) + 3 (3 x - 2) = 0$

$\Rightarrow (4 x + 3) (3 x - 2) = 0$

Therefore, $f (x) = (x + 2) (12 x^{2} + x - 6) = (x + 2) (3 x - 2) (4 x + 3)$

Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Logarithmic and Exponential Functions, Exercise 10: END-OF-CHAPTER REVIEW EXERCISE 2

Sue Pemberton Mathematics Solutions for Exercise - Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Logarithmic and Exponential Functions, Exercise 10: END-OF-CHAPTER REVIEW EXERCISE 2

Questions from Sue Pemberton, Julianne Hughes and, Julian Gilbey Solutions for Chapter: Logarithmic and Exponential Functions, Exercise 10: END-OF-CHAPTER REVIEW EXERCISE 2 with Hints & Solutions

Learn and Prepare for any exam you want !