Question

The zeroes of the polynomial $P(x) = \left( {x - 6} \right)\left( {x - 5} \right)$ are:
A. -6, -5
B. -6, 5
C. 6, -5
D. 6, 5

Answer 1

Hint: To solve the question given above first, we will find out what is a polynomial, what kind of polynomial it is, and how many roots it will have. Then, we will put $P(x)=0$ to get the zeroes of the equation obtained. We will find the zeros of the equation by completing the square method. In this method, we will try to form squares on both sides of the equation.

Complete step-by-step answer:
Before we solve the question given above, we must know what polynomial is and what kind of polynomial it is. A polynomial is an expression consisting of variables and coefficients that involves only the operation of addition, subtraction, multiplication and non-negative integer exponents of variables. Now, the maximum power of x in the given polynomial is 2. So, it is a quadratic polynomial. As it is a quadratic polynomial, so, we will get two roots when we equate P(x) to zero. Now, we will expand the polynomial P(x).
\[\begin{align}
  & P(x)=\left( x-6 \right)\left( x-5 \right) \\
 & P(x)={{x}^{2}}-5x-6x+30 \\
 & P(x)={{x}^{2}}-11x+30 \\
\end{align}\]
Now, to obtain the zeroes, we equate P(x) to zeroes. Thus, we will get,
\[\begin{align}
  & P(x)={{x}^{2}}-11x+30=0 \\
 & \Rightarrow {{x}^{2}}-11x+30=0 \\
\end{align}\]
Now, we will solve the above quadratic equation by completing the square method. In this method, we try to form a square on both sides of the equation. For making perfect square of given quadratic equation we add and subtract the square of half of the coefficient of x:
\[\begin{align}
  & \Rightarrow {{x}^{2}}-11x+30=0 \\
 & \Rightarrow {{x}^{2}}-2\times x\times \dfrac{11}{2}+{{\left( \dfrac{11}{2} \right)}^{2}}-{{\left( \dfrac{11}{2} \right)}^{2}}+30=0 \\
 & \Rightarrow \left[ {{x}^{2}}-11x+{{\left( \dfrac{11}{2} \right)}^{2}} \right]={{\left( \dfrac{11}{2} \right)}^{2}}-30 \\
 & \Rightarrow {{\left( x-\dfrac{11}{2} \right)}^{2}}=\dfrac{121}{4}-30 \\
 & \Rightarrow {{\left( x-\dfrac{11}{2} \right)}^{2}}=\dfrac{1}{4} \\
 & \Rightarrow {{\left( x-\dfrac{11}{2} \right)}^{2}}={{\left( \dfrac{1}{2} \right)}^{2}} \\
\end{align}\]
Now, we will take square root on both sides. Thus, we will get:
\[\begin{align}
  & \left( x-\dfrac{11}{2} \right)=\pm \dfrac{1}{2} \\
 & \Rightarrow x-\dfrac{11}{2}=\dfrac{1}{2}\text{ and }x-\dfrac{11}{2}=-\dfrac{1}{2}\text{ } \\
 & \Rightarrow \text{x=}\dfrac{11}{2}+\dfrac{1}{2}\text{ and x=}\dfrac{11}{2}-\dfrac{1}{2} \\
 & \Rightarrow x=6\text{ and x=5} \\
\end{align}\]
Therefore, the zeroes of P(x) are x=6 and x=5. Hence, option D is correct.

Note: The zeroes of the quadratic equation obtained can also be find out by using the quadratic formula. If the quadratic equation is of the form $a{{x}^{2}}+bx+c=0$ then, we can say that:
\[\begin{align}
  & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{-\left( -11 \right)\pm \sqrt{{{\left( -11 \right)}^{2}}-4\left( 1 \right)\left( 30 \right)}}{2\left( 1 \right)} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow x=\dfrac{11\pm \sqrt{121-120}}{2} \\
 & \Rightarrow x=\dfrac{11\pm \sqrt{1}}{2} \\
 & \Rightarrow x=\dfrac{11\pm 1}{2} \\
 & \Rightarrow x=\dfrac{12}{2}\text{ or x=}\dfrac{10}{2} \\
 & \Rightarrow x=5\text{ and x=6} \\
\end{align}\]