Question

How do you factor the expressions ${{y}^{2}}-11y+30$?

Answer 1

Hint: In this problem we need to calculate the factors of the given equation which is a quadratic equation. For a quadratic equation which is in the form of $a{{x}^{2}}+bx+c$. We will split the middle term $bx$ as $bx={{b}_{1}}x+{{b}_{2}}x$ such that ${{b}_{1}}{{b}_{2}}=ac$. So, we will first compare the given equation with the standard form of the quadratic equation and write the value of $a$, $b$, $c$. After that we will select any two of values from the factors of the value $ac$ to split the middle term. Now we will take appropriate terms as common from the equation to get the required result.

Complete step by step solution:
Given that, ${{y}^{2}}-11y+30$.
Comparing the above equation with the standard form of quadratic equation, then we will get
$a=1$, $b=-11$, $c=30$.
The value of $ac$ will be $ac=1\times 30=30$ and the factors of the value $30$ are $1$, $2$, $3$, $5$, $6$, $10$, $15$, $30$. From these factors we can write that
$\begin{align}
  & -6-5=-11 \\
 & -6\times \left( -5 \right)=30 \\
\end{align}$
So, splitting the middle term $-11y$ as $-6y-5y$, then the given equation is modified as
$\Rightarrow {{y}^{2}}-11y+30={{y}^{2}}-6y-5y+30$
Taking $y$ as common from the first two terms and $-5$ as common from the last two terms, then we will get
$\Rightarrow {{y}^{2}}-11y+30=y\left( y-6 \right)-5\left( y-6 \right)$
Again, taking $y-6$ as common from the above equation, then we will have
$\Rightarrow {{y}^{2}}-11y+30=\left( y-6 \right)\left( y-5 \right)$
Hence the roots of the given equation ${{y}^{2}}-11y+30$ are $y-6$, $y-5$.

Note: We can also follow another method to solve this problem. We will first calculate the roots of the given equation by using the quadratic formula which is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. If we have the roots of the quadratic equation as $\alpha $, $\beta $ then the factors will be $x-\alpha $, $x-\beta $.