Question

How do you factor the $y=2{{x}^{2}}+11x+12?$

Answer 1

Hint: To find the factor of given trinomial which is basically a quadratic polynomial we will first take 2 as common and make coefficient of ${{x}^{2}}$ as 1 and then we will find the root of the given quadratic polynomial obtained by using the formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ , let us say that root is $\alpha ,\beta $ then the factor of the given polynomial obtained after taking common 2 will be $\left( x-\alpha \right)\left( x-\beta \right)$ .

Complete step by step answer:
We will start solving the given polynomial by first recalling the Euclidean division of polynomials .
We know that when we divide a divided polynomial $p\left( x \right)$ with degree $n$ by some divisor polynomial $d\left( x \right)$ with degree $m\le n$ then we get the quotient polynomial $q\left( x \right)$ of degree $n-m$ and the remainder polynomial as $r\left( x \right)$ .We use Euclidean division formula and can write as
$p\left( x \right)=d\left( x \right)q\left( x \right)+r\left( x \right)$
We also know that if the remainder polynomial is zero then we call $d\left( x \right),q\left( x \right)$ factor polynomial of $p\left( x \right)$ or simply factors of $p\left( x \right)$ . If ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right).......{{p}_{k}}\left( x \right)$ are $k$ factors of $p\left( x \right)$ then we say ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right).......{{p}_{k}}\left( x \right)$ is factorized completely if none of the factors ${{p}_{1}}\left( x \right),{{p}_{2}}\left( x \right).......{{p}_{k}}\left( x \right)$ can be factorized further.
So, to find the factor of the given trinomial $2{{x}^{2}}+11x+12$, we will at first take 2 as common factor and then we can write $2{{x}^{2}}+11x+12$ as:
\[\begin{align}
  & \Rightarrow 2{{x}^{2}}+11x+12 \\
 & \Rightarrow 2\left( {{x}^{2}}+\dfrac{11}{2}x+6 \right) \\
\end{align}\]
Now, to further factorize will first find the root of \[\left( {{x}^{2}}+\dfrac{11}{2}x+6 \right)=0\], and if the root of the equation is $\alpha ,\beta $, then the factor of the given polynomial obtained after taking common 2 will be $\left( x-\alpha \right)\left( x-\beta \right)$ .
Now, for the general quadratic equation $a{{x}^{2}}+bx+c=0$, we know that that its roots are given by:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So, for the quadratic equation\[\left( {{x}^{2}}+\dfrac{11}{2}x+6 \right)=0\], roots are equal to:
$\Rightarrow x=\dfrac{-\dfrac{11}{2}\pm \sqrt{{{\left( \dfrac{11}{2} \right)}^{2}}-4\times 1\times 6}}{2\times 1}$
$\Rightarrow x=\dfrac{-\dfrac{11}{2}\pm \sqrt{\dfrac{121}{4}-24}}{2\times 1}$
$\Rightarrow x=\dfrac{-\dfrac{11}{2}\pm \sqrt{\dfrac{121-24\times 4}{4}}}{2\times 1}=\dfrac{-\dfrac{11}{2}\pm \sqrt{\dfrac{121-96}{4}}}{2}$
$\Rightarrow x=\dfrac{-\dfrac{11}{2}\pm \sqrt{\dfrac{25}{4}}}{2}$
$\Rightarrow x=\dfrac{-\dfrac{11}{2}\pm \dfrac{5}{2}}{2}$
Now, we will take 2 as LCM on numerator then we will get:
$\Rightarrow x=\dfrac{-11\pm 5}{2\times 2}$
$\Rightarrow x=\dfrac{-11+5}{4},\dfrac{-11-5}{4}$
$\Rightarrow x=\dfrac{6}{4},\dfrac{-16}{4}$
$\therefore x=\dfrac{3}{2},-4$
Hence, $\left( x-\dfrac{3}{2} \right),\left( x-\left( -4 \right) \right)$ are the factor of the polynomial \[\left( {{x}^{2}}+\dfrac{11}{2}x+6 \right)\].
So, we can write $2{{x}^{2}}+10x+25=2\left( x-\dfrac{3}{2} \right)\left( x+4 \right)$
Hence, $2,\left( x-\dfrac{3}{2} \right),\left( x+4 \right)$ are the factors of the trinomial $2{{x}^{2}}+11x+12$
This is our required solution.

Note: We note that the highest power on the variable is called degree of the polynomial. If degree is 1 we call the polynomial a linear polynomial. If the polynomial is a single term then we call the polynomial a monomial and if it has two terms it is called a binomial. And if the factor has three terms then it is called a trinomial. Hence, $2{{x}^{2}}+11x+12$ is a trinomial. Also, note that when we equate factor with zero then we will get the root or we can say the solution of that polynomial.