Question

How do you factor $2{{x}^{2}}-11x+5=0$?

Answer 1

We use both grouping method and vanishing method to find the factor of the problem. We take common terms out to form the multiplied forms. In the case of the vanishing method, we use the value of x which gives the polynomial value 0.

Complete step-by-step solution:
We apply the middle-term factoring or grouping to factorise the polynomial.
Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number.
In the case of $2{{x}^{2}}-11x+5$, we break the middle term $-11x$ into two parts of $-10x$ and $-x$.
So, $2{{x}^{2}}-11x+5=2{{x}^{2}}-10x-x+5$. We have one condition to check if the grouping is possible or not. If we order the individual elements of the polynomial according to their power of variables, then the multiple of end terms will be equal to the multiple of middle terms.
Here multiplication for both cases gives $10{{x}^{2}}$. The grouping will be done for $2{{x}^{2}}-10x$ and $-x+5$. We try to take the common numbers out.
For $2{{x}^{2}}-10x$, we take $2x$ and get $2x\left( x-5 \right)$.
For $-x+5$, we take $-1$ and get $-\left( x-5 \right)$.
The equation becomes $2{{x}^{2}}-11x+5=2{{x}^{2}}-10x-x+5=2x\left( x-5 \right)-\left( x-5 \right)$.
Both the terms have $\left( x-5 \right)$ in common. We take that term again and get
$\begin{aligned} & 2{{x}^{2}}-11x+5 \\\ & =2x\left( x-5 \right)-\left( x-5 \right) \\\ & =\left( x-5 \right)\left( 2x-1 \right) \\\ \end{aligned}$
Therefore, the factorisation of $2{{x}^{2}}-11x+5$ is $\left( x-5 \right)\left( 2x-1 \right)$.

Note: We find the value of x for which the function $f\left( x \right)=2{{x}^{2}}-11x+5=0$. We can see $f\left( 5 \right)=2\times {{5}^{2}}-11\times 5+5=50-55+5=0$. So, the root of the $f\left( x \right)=2{{x}^{2}}-11x+5$ will be the function $\left( x-5 \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
Now, $f\left( x \right)=2{{x}^{2}}-11x+5=\left( x-5 \right)\left( 2x-1 \right)$. We can also do this for $\left( 2x-1 \right)$.