Question

How do you express $f\left( x \right)={{x}^{3}}+4{{x}^{2}}-5x-14$ as a product of linear factors?

Answer 1

Hint: In this question we have been asked to express the given function $f\left( x \right)={{x}^{3}}+4{{x}^{2}}-5x-14$ as a product of linear factors. For that first we will replace $x$ with some number and verify $f\left( x \right)=0$ or not then if it is we can say the number is a factor of the given function.

Complete step by step solution:
Now considering from the question we have been asked to express the given function $f\left( x \right)={{x}^{3}}+4{{x}^{2}}-5x-14$ as a product of linear factors.
From the basic concepts we know that we can say that a number is a factor of the given function if $x$ satisfies it that is $f\left( x \right)=0$ .
Let us verify for $x=2$ . By replacing $x$ with $2$ we will have
$\begin{align}
  & \Rightarrow f\left( 2 \right)={{\left( 2 \right)}^{3}}+4{{\left( 2 \right)}^{2}}-5\left( 2 \right)-14 \\
 & \Rightarrow f\left( 2 \right)=8+16-10-14 \\
 & \Rightarrow f\left( 2 \right)=0 \\
\end{align}$
Hence $2$ is a factor.
Now we can express this function as $\Rightarrow f\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}+6x+7 \right)$ .
We know that we can factorize any quadratic expression in the form of $a{{x}^{2}}+bx+c$ using the formula for finding its roots which is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . If ${{x}_{1}},{{x}_{2}}$ are roots of the quadratic expression then its factors will be $\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)$ .
We can factorize ${{x}^{2}}+6x+7$using the concept. The roots of the expression are given by
$\begin{align}
  & \Rightarrow x=\dfrac{-6\pm \sqrt{{{6}^{2}}-4\left( 1 \right)\left( 7 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{-6\pm \sqrt{36-28}}{2} \\
 & \Rightarrow x=\dfrac{-6\pm \sqrt{8}}{2} \\
 & \Rightarrow x=-3\pm \sqrt{2} \\
\end{align}$
Therefore the factors of the expression will be $\left( x+3\mp \sqrt{2} \right)$ .
Therefore we can conclude that the expression $f\left( x \right)={{x}^{3}}+4{{x}^{2}}-5x-14$ can be expressed as a product of linear factors as $f\left( x \right)=\left( x-2 \right)\left( x+3-\sqrt{2} \right)\left( x+3+\sqrt{2} \right)$.

Note: While answering this question we should be sure with our concepts that we apply and calculations that we perform during the process of solving. Similarly we have many methods for factoring different expressions. For example the traditional method for factorising a quadratic expression is given as we need to express the coefficient of $x$ as the sum of the factors of the product of the constant term and coefficient of ${{x}^{2}}$ .We can consider for example ${{x}^{2}}+5x+4$ can be done as ${{x}^{2}}+5x+4={{x}^{2}}+x+4x+4=x\left( x+1 \right)+4\left( x+1 \right)=\left( x+4 \right)\left( x+1 \right)$ .