Question

How do you simplify $\left( x-2 \right)\left( x+4 \right)\left( x+6 \right)$ ?

Answer 1

Hint: We are given an expression in one variable which has been written as the multiplication of three terms, each being a linear equation. Thus, we will open these brackets one by one and multiply the terms in it with those inside the other brackets. Further, we shall use basic algebra to combine the like terms and present the equation in the simplest form possible.

Complete step-by-step solution:
The given equation consists of three brackets each one containing a linear equation in x-variable along with a constant term only. Therefore, from our prior knowledge of polynomials, we understand that this equation will be transformed into a three degree equation in x-variable.
Given that $\left( x-2 \right)\left( x+4 \right)\left( x+6 \right)$, we shall first open the brackets of $x-2$ and $x+4$ to multiply these terms with each other to obtain a quadratic equation in x-variable.
$\Rightarrow \left( {{x}^{2}}-2x+4x-8\right)\left( x+6 \right)$
Combining the terms and subtracting 2x from 4x, we get
$\Rightarrow \left({{x}^{2}}+2x-8\right)\left( x+6 \right)$
Now, we shall open the brackets of $x+6$ and multiply $x$ and 6 separately with the obtained quadratic equation.
$\begin{align}
  & \Rightarrow x\left( {{x}^{2}}+2x-8 \right)+6\left( {{x}^{2}}+2x-8 \right) \\
 & \Rightarrow {{x}^{3}}+2{{x}^{2}}-8x+6{{x}^{2}}+12x-48 \\
\end{align}$
Combining the terms of ${{x}^{2}}$, $x$ and constant terms using basic algebra, we get
$\Rightarrow {{x}^{3}}+8{{x}^{2}}+4x-48$
Therefore, the simplified form of $\left( x-2 \right)\left( x+4 \right)\left( x+6 \right)$ is given as ${{x}^{3}}+8{{x}^{2}}+4x-48$.

Note: The given equation in the question, $\left( x-2 \right)\left( x+4 \right)\left( x+6 \right)$ also signifies that the roots of this equation are 2, -4 and -6. This can be obtained by equating the expression with zero and then separately taking each linear equation in each bracket as equal to zero. On further transposing the constant term on the right-hand side, we shall get the value of x.