Question

How do you factor completely $4{{x}^{3}}+28{{x}^{2}}+7x+49$ ?

Answer 1

Hint: Now to solve the given expression we will first simplify the given expression by taking $4{{x}^{2}}$ common from the first two terms and $7$ common from the last two terms. Hence we will find the first factor of the given expression. Now we will solve for the roots of the quadratic by using the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Now we know that the factor of the quadratic expression is $x-\alpha $ and $x-\beta $ where $\alpha $ and $\beta $ are the roots of the expression.

Complete step by step solution:
Now consider the given expression $4{{x}^{3}}+28{{x}^{2}}+7x+49$
Now to simplify the expression we will first take $4{{x}^{2}}$ common from the first term and the second terms. Hence we get,
$\Rightarrow 4{{x}^{2}}\left( x+7 \right)+7x+49$
Now we will take 7 common from the last two terms. Hence we get the expression as,
$\Rightarrow 4{{x}^{2}}\left( x+7 \right)+7\left( x+7 \right)$
Now again taking $\left( x+7 \right)$ common from the first two terms we get,
$\Rightarrow \left( 4{{x}^{2}}+7 \right)\left( x+7 \right)$
Now consider the expression $4{{x}^{2}}+7$ .
Now this is a quadratic expression of the form $a{{x}^{2}}+bx+c$ .
Now first let us find the roots of the expression.
Hence consider $4{{x}^{2}}+7=0$
Now taking 7 on RHS and dividing the equation by 4 we get,
$\Rightarrow {{x}^{2}}=\dfrac{-7}{4}$
Now taking square root on both sides we get,
$\Rightarrow x=\pm \dfrac{\sqrt{7}i}{2}$ where $i=\sqrt{-1}$
Hence we get the roots of the expression are $-\dfrac{i\sqrt{7}}{2}$ and $\dfrac{i\sqrt{7}}{2}$
Now we know that if $\alpha $ and $\beta $ are the roots of the expression then $x-\alpha $ and $x-\beta $ are the factors of the expression.
Hence we get,
$\Rightarrow 4{{x}^{2}}+7=\left( x+\dfrac{i\sqrt{7}}{2} \right)\left( x-\dfrac{i\sqrt{7}}{2} \right)$
Now substituting this value in $\left( 4{{x}^{2}}+7 \right)\left( x+7 \right)$ we get,
$\Rightarrow \left( x+\dfrac{i\sqrt{7}}{2} \right)\left( x-\dfrac{i\sqrt{7}}{2} \right)\left( x+7 \right)$
Hence we have the factors of the given expression.

Note: Now that we here we get complex roots of the quadratic. Now if the discriminant of the quadratic which is defined by $D={{b}^{2}}-4ac$ is negative then we get complex roots. Remember that the complex roots always occur in pair of conjugates.