Question

If the zeroes of the polynomial $P\left(x\right)=a{x}^2+bx+c$ are 1 and 2 , then find the value of a, b and c.

• A. a = 1, b = -3 , c = 1
• B. a = 1 , b = -3, c = 2
• C. a = 3, b = -1, c = 1
• D. All of these

Answer 1

The correct option is B

a = 1 , b = -3, c = 2

If the factors of the polynomial equation $P_1\left(x\right)=x^2+6x+5$ are (x-a) & (x-b), then $P_1\left(x\right)=(x–a)(x–b)$
$P_1\left(x\right)=(x–a)(x-b)=x^2+6x+5$.
By comparing the coefficients of $x^2$, $x$ and the constant term, we can find the value of ‘a’ and ‘b’.
Now in the above question the zeroes of $P\left(x\right)=a{x}^2+bx+c$ are 1 and 2.
Therefore $P\left(x\right)=(x–1)(x-2)=x^2–3x+2$.
Now comparing the coefficients and the constant term of $x^2–3x+2$ and $a{x}^2+bx+c$ we get a = 1 , b = -3, c = 2.