Question

How do you factor \[y={{x}^{3}}+6{{x}^{2}}+8x\]?

Answer 1

Hint: For answering this question we will use factorization. Factorization is the process of deriving factors of a number which divides the given number evenly. For this question we will bring \[x\] common in the expression and assume the remaining quadratic term as \[f\left( x \right)\]and will split the constant term in \[f\left( x \right)\] and we will use the sum product pattern of splitting and then we will simplify the equation.

Complete step by step solution:
Now considering from the question we have an expression \[{{x}^{3}}+6{{x}^{2}}+8x\] for which we need to derive the factors.
We can factor the \[{{x}^{3}}+6{{x}^{2}}+8x\] by below method:
Given equation is in the form of \[x\left( a{{x}^{2}}+bx+c \right)=0\].
So, it can be written as \[x\left( {{x}^{2}}+6x+8 \right)\]. Let us assume \[{{x}^{2}}+6x+8\] as \[f\left( x \right)\].
First we have to divide the constant term in the \[f\left( x \right)\] which is \[8\] into the product of the two numbers and must make sure that the sum of the two numbers must be equal to the coefficient of\[x\]in \[f\left( x \right)\].
Now, the constant term \[8\] can be split into the product of the two numbers in two ways those are \[8\times 1,4\times 2\].
But here we have to take the splitting as \[4\times 2\] as the sum of \[4\] and 2 must be equal to the coefficient of \[x\].
So, 8 can be split into products of \[2\]and \[4\].
Their sum is also equal to \[6\]which is equal to the coefficient of \[x\].
So, the given question can be factored as follows.
\[\Rightarrow x\left( {{x}^{2}}+6x+8 \right)\]
\[\Rightarrow x\left( {{x}^{2}}+4x+2x+8 \right)\]
\[\Rightarrow x\left[ x\left( x+4 \right)+2\left( x+4 \right) \right]\]
\[\Rightarrow x\left[ \left( x+2 \right)\left( x+4 \right) \right]\]
\[\Rightarrow x\left( x+2 \right)\left( x+4 \right)\]

Note: During answering questions of this type we should be sure with our calculations. we can also use the formulae for \[f\left( x \right)\] in \[xf\left( x \right)\]to obtaining the roots by comparing with quadratic equation $a{{x}^{2}}+bx+c=0$ and solve the question. So, the roots of the quadratic equation given as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ then if the two solutions are $p,q$ then the factors will be $\left( x-p \right)\left( x-q \right)$ . For \[f\left( x \right)\] the roots are \[\dfrac{-6\pm \sqrt{36-4\left( 8 \right)}}{2}=\dfrac{-6\pm 2}{2}=-4,-2\] then the factors will be \[\left( x+2 \right),\left( x+4 \right)\] for \[f\left( x \right)\]. Therefore, the factor for the question will be \[x,\left( x+2 \right),\left( x+4 \right)\].