Question

If both $\left( {x - 2} \right)$ and $\left( {x - \dfrac{1}{2}} \right)$ are factors of $p{x^2} + 5x + r$, show that $p = r$.

Answer 1

Hint: We have 2 factors of the expression. From the 2 factors, we can find 2 roots of the polynomial. With the roots, we can get 2 linear equations in terms of $p$ and $r$. By equating the 2 linear equations, we get the required relation.

Complete step by step answer:

We have the polynomial $p{x^2} + 5x + r$with factors $\left( {x - 2} \right)$and $\left( {x - \dfrac{1}{2}} \right)$.
From the factors, we get, $x = 2$and $x = \dfrac{1}{2}$ are the solutions of the equation.
When $x = 2$, the expression becomes,
\[
  p \times {2^2} + 5 \times 2 + r = 0 \\
   \Rightarrow 4p + 10 + r = 0 \\
 \]
When $x = \dfrac{1}{2}$, the expression becomes,
\[
  p \times {\left( {\dfrac{1}{2}} \right)^2} + 5 \times \dfrac{1}{2} + r = 0 \\
   \Rightarrow \dfrac{p}{4} + \dfrac{5}{2} + r = 0 \\
   \Rightarrow p + 10 + 4r = 0 \\
 \]
By equating both the equations, we get,
\[4p + 10 + r = p + 10 + 4r\]
On subtracting with 10 on both sides of the equation, we get.
\[4p + r = p + 4r\]
Subtracting both sides with p and r,
\[3p = 3r\]
On dividing both sides with 3, we get
\[p = r\]
Hence, we get the required relation.

Note: The concept used here is factors and solutions of polynomials. The values of the variable at which the polynomial attains value zero is called roots of the polynomial. If $x - a$ is a factor of a polynomial, the $x = a$ is a root of the polynomial. Then the value of the polynomial at $x = a$ will be zero. The methods of simple algebra are used to derive the required relation. We can also find the value of p and r using the relation in any of the linear equations we found out.