Question

If both $(x-2)$ and $(x-\frac{1}{2})$ are factors of $(a{x}^2+5x+b)$, show that $a=b$

Answer 1

The Factor Theorem states that if $a$ is the root of any polynomial $p\left(x\right)$ that is if $p\left(a\right)=0$, then $(x-a)$ is the factor of the polynomial $p\left(x\right)$.

Let $p\left(x\right)=a{x}^2+5x+b$. It is given that $(x-2)$ and $(x-\frac{1}{2})$, therefore, by factor theorem $p\left(2\right)=0$ and $p\left(\frac{1}{2}\right)=0$. Let us first find $p\left(2\right)$ and $p\left(\frac{1}{2}\right)$ as follows:

$p\left(2\right)=(a\times 2^2)+(5\times 2)+b=(a\times 4)+10+b=4a+b+10p\left(\frac{1}{2}\right)=a{\left(\frac{1}{2}\right)}^2+(5\times \frac{1}{2})+b=(a\times \frac{1}{4})+\frac{5}{2}+b=\frac{a}{4}+b+\frac{5}{2}$

Now equate $p\left(2\right)=0$ and $p\left(\frac{1}{2}\right)=0$ as shown below:

$4a+b+10=0\Rightarrow 4a+b=-10.......\left(1\right)\frac{a}{4}+b+\frac{5}{2}=0\Rightarrow \frac{a+4b+10}{4}=0\Rightarrow a+4b+10=0\Rightarrow a+4b=-10.......\left(2\right)$

Now equate equation 1 and 2:

$4a+b=a+4b\Rightarrow 4a-a=4b-b\Rightarrow 3a=3b\Rightarrow a=b$

Hence, $a=b$