Question

How do you factor $y = 2{x^2} - 9x - 18?$

Answer 1

Hint: Here we need to factorize the given polynomial $2{x^2} - 9x - 18$. 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 $y = 2{x^2} - 9x - 18$
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 $a = 2$, $b = - 9$ and $c = - 18.$
We split the middle term $ - 9x$ as, $ - 9x = 3x - 12x$.
Note that their product is,
 $a \cdot c = 2 \times ( - 18)$
$ \Rightarrow a \cdot c = - 36$
$ \Rightarrow - 36 = 3 \times ( - 12)$.
Note that their sum is,
$b = - 9$
$ \Rightarrow - 9 = 3 - 12$.
Hence the equation $2{x^2} - 9x - 18$can be written as,
$ \Rightarrow 2{x^2} + 3x - 12x - 18$
Factor out the greatest common factor from each group, we get,
$ \Rightarrow x(2x + 3) - 6(2x + 3)$
Now factor the polynomial by factoring out the greatest common factor $2x + 3$, we get,
$ \Rightarrow (2x + 3)(x - 6)$

Hence the factorization of the equation $y = 2{x^2} - 9x - 18$ is $(2x + 3)(x - 6)$.

Note :
Alternative method :
Given a quadratic equation $y = 2{x^2} - 9x - 18$.
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 $a = 2$, $b = - 9$ and $c = - 18.$
Substituting these values in the formula we get,
$ \Rightarrow x = \dfrac{{ - ( - 9) \pm \sqrt {{{( - 9)}^2} - 4 \times 2 \times ( - 18)} }}{{2 \times 2}}$
$ \Rightarrow x = \dfrac{{9 \pm \sqrt {81 - 8 \times ( - 18)} }}{4}$
$ \Rightarrow x = \dfrac{{9 \pm \sqrt {81 + 144} }}{4}$
$ \Rightarrow x = \dfrac{{9 \pm \sqrt {225} }}{4}$
We know that $\sqrt {225} = 15$.
$ \Rightarrow x = \dfrac{{9 \pm 15}}{4}$
Hence we get two roots given by,
${x_1} = \dfrac{{9 + 15}}{4}$ and ${x_2} = \dfrac{{9 - 15}}{4}$
$ \Rightarrow {x_1} = \dfrac{{24}}{4}$ and ${x_2} = - \dfrac{6}{4}$
$ \Rightarrow {x_1} = 6$ and ${x_2} = - \dfrac{3}{2}$
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 - \left( { - \dfrac{3}{2}} \right)} \right)$
$ \Rightarrow \left( {x - 6} \right)\left( {x + \dfrac{3}{2}} \right)$
$ \Rightarrow \left( {x - 6} \right)\left( {2x + 3} \right)$
Hence the factorization of the equation $y = 2{x^2} - 9x - 18$ is given by $(2x + 3)(x - 6)$.