If the polynomial P(x) = x^2 + ax + b has fac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $P(x) = x^2 + ax + b = (x - a)(x - b)$.Expanding the right side: $x^2 - (a + b)x + ab = x^2 + ax + b$.Comparing coefficients:$a = -(a + b)$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given $P(x) = x^2 + ax + b = (x - a)(x - b)$.Expanding the right side: $x^2 - (a + b)x + ab = x^2 + ax + b$.Comparing coefficients:$a = -(a + b)$ $\Rightarrow 2a + b = 0$ $\Rightarrow b = -2a$.$b = ab$.Case $1$: If $b 
eq 0$,then $a = 1$. Substituting into $b = -2a$,we get $b = -2$.Then $P(x) = x^2 + x - 2$. $P(2) = 2^2 + 2 - 2 = 4$.Case $2$: If $b = 0$,then $2a + 0 = 0 \Rightarrow a = 0$.Then $P(x) = x^2$. $P(2) = 2^2 = 4$.In both cases,$P(2) = 4$.

If the polynomial $P(x) = x^2 + ax + b$ has factors $(x - a)(x - b)$,where $a, b \in R$,then the value of $P(2)$ is:

Similar Questions

The condition that $x^3 - p x^2 + q x - r = 0$ may have two of its roots equal to each other but of opposite sign is

If one root of the quadratic equation $ax^2 + bx + c = 0$ is equal to the $n$th power of the other,then $(ac^n)^{1/(n+1)} + (a^nc)^{1/(n+1)} =$

If one real root of the quadratic equation $81x^2 + kx + 256 = 0$ is the cube of the other root,then a value of $k$ is

If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$,then the value of $\alpha^6+\beta^6$ is

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 2(m^2 + 1)x + m^4 + m^2 + 1 = 0$,then find the value of $\alpha^2 + \beta^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module