Question

If $\left( x+p \right)$ is a factor of the polynomial $2{{x}^{2}}+2px+5x+10$. Find the value of ‘p’.

Answer 1

Hint: We have a polynomial $2{{x}^{2}}+2px+5x+10$ which has a factor $\left( x+p \right)$. We use the fact that factor divides the polynomial and gives a zero remainder. So, on substituting x = - p in the polynomial we will get value to be zero. By substituting -p in place of ‘x’ we get an equation to solve. On solving the equation, we get the final value of ‘p’.

Complete step by step answer:
Given that we have a polynomial $2{{x}^{2}}+2px+5x+10$ which has a factor $\left( x+p \right)$. We need to find the value of ‘p’.
Before finding the value of ‘p’, we first get to know about the factor of the polynomial.
If $\left( x+a \right)$ is factor of the polynomial $p\left( x \right)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+......+{{a}_{1}}x+{{a}_{0}}$, then the value of $P\left( -a \right)$ is equal to zero. In simple words factor gives a remainder zero on dividing the polynomial.
For finding $P\left( -a \right)$, we substitute ‘-a’ in place of ‘x’ in the polynomial.
Let us the polynomial $Q\left( x \right)=2{{x}^{2}}+2px+5x+10$ and $\left( x+p \right)$ is the factor of $Q\left( x \right)$. So, we get $Q\left( -p \right)=0$.
$Q\left( -p \right)=0$.
$2{{\left( -p \right)}^{2}}+2.p.\left( -p \right)+5\left( -p \right)+10=0$.
$2\left( {{p}^{2}} \right)-2{{p}^{2}}-5p+10=0$.
$2{{p}^{2}}-2{{p}^{2}}-5p+10=0$.
$-5p+10=0$.
\[5p=10\].
$p=\dfrac{10}{5}$.
$p=2$.

∴ The value of p is 2.

Note: We should not confuse and substitute ‘p’ in place of ‘-p’ in the polynomial as it is the most common mistake any person can make. While calculating the value of ‘p’, we should check the sign convention properly while multiplying and adding two or more terms. The other type of problems we get here is finding the quotient and remainder when the polynomial is divided by a factor.