Question

How do you factor $y = {x^2} + 2x - 48?$

Answer 1

Hint: Here we need to factorize the given polynomial ${x^2} + 2x - 48$. Note that the degree of the polynomial is 2, so it is a quadratic polynomial. This polynomial is of the form $a{x^2} + bx + c$. So we rewrite the middle term as a sum of terms whose product is $a \cdot c$ and whose sum is b. Therefore, substitute the values of a, b, c and solve the given problem by splitting the middle term.

Complete step by step solution:
Given an equation $y = {x^2} + 2x - 48$
It is mentioned that we need to factor the given polynomial.
Note that the given equation is a quadratic equation, so we factor it by splitting the middle term.
Consider an equation of the form $a{x^2} + bx + c$, where a, b, c are any real numbers.
We rewrite the middle term as a sum of two terms in such a way that their product is $a \cdot c$ and their sum is b.
In the given equation we have $a = 1$, $b = 2$ and $c = - 48.$
We split the middle term $2x$ as, $2x = 8x - 6x$.
Note that their product is,
 $a \cdot c = 1 \times ( - 48)$
$ \Rightarrow a \cdot c = - 48$
$ \Rightarrow - 48 = 8 \times ( - 6)$.
Note that their sum is,
$b = 2$
$ \Rightarrow 2 = 8 - 6$.
Hence the equation ${x^2} + 2x - 48$ can be written as,
$ \Rightarrow {x^2} + 8x - 6x - 48$
Factor out the greatest common factor from each group, we get,
$ \Rightarrow x(x + 8) - 6(x + 8)$
Now factor the polynomial by factoring out the greatest common factor $x + 8$, we get,
$ \Rightarrow (x + 8)(x - 6)$

Hence the factorization of the equation $y = {x^2} + 2x - 48$ is $(x + 8)(x - 6)$.

Note :
Alternative method :
Given a quadratic equation $y = {x^2} + 2x - 48$.
Here we find the roots of the given equation and then from the roots we try to find out the factors.
This equation is of the form of $a{x^2} + bx + c$. We find the roots using the formula,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Here $a = 1$, $b = 2$ and $c = - 48.$
Substituting these values in the formula we get,
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{{(2)}^2} - 4 \times 1 \times ( - 48)} }}{{2 \times 1}}$
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 - 4 \times ( - 48)} }}{2}$
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 + 192} }}{2}$
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {196} }}{2}$
We know that $\sqrt {196} = 14$.
$ \Rightarrow x = \dfrac{{ - 2 \pm 14}}{2}$
Hence we get two roots given by,
${x_1} = \dfrac{{ - 2 + 14}}{2}$and ${x_2} = \dfrac{{ - 2 - 14}}{2}$
$ \Rightarrow {x_1} = \dfrac{{12}}{2}$and ${x_2} = - \dfrac{{16}}{2}$
$ \Rightarrow {x_1} = 6$ and ${x_2} = - 8$
So, the factors for the quadratic equation when the roots are given is found using the formula,
$(x - {x_1})(x - {x_2})$
Substituting the values of ${x_1}$ and ${x_2}$ we get the required factors.
$ \Rightarrow \left( {x - 6} \right)\left( {x - ( - 8)} \right)$
$ \Rightarrow \left( {x - 6} \right)\left( {x + 8} \right)$
Hence the factorization of the equation $y = {x^2} + 2x - 48$ is given by $\left( {x - 6} \right)\left( {x + 8} \right)$.