If (x + 1) is a factor of {x^4} - (p - 3){x^3 | vedclass

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

Question

(A) Let $f(x) = {x^4} - (p - 3){x^3} - (3p - 5){x^2} + (2p - 7)x + 6$.Since $(x + 1)$ is a factor of $f(x)$,by the Factor Theorem,$f(-1) = 0$.Substitu

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let $f(x) = {x^4} - (p - 3){x^3} - (3p - 5){x^2} + (2p - 7)x + 6$.Since $(x + 1)$ is a factor of $f(x)$,by the Factor Theorem,$f(-1) = 0$.Substituting $x = -1$ into the polynomial:$(-1)^4 - (p - 3)(-1)^3 - (3p - 5)(-1)^2 + (2p - 7)(-1) + 6 = 0$$1 - (p - 3)(-1) - (3p - 5)(1) - (2p - 7) + 6 = 0$$1 + (p - 3) - (3p - 5) - 2p + 7 + 6 = 0$$1 + p - 3 - 3p + 5 - 2p + 7 + 6 = 0$Combining like terms:$(1 - 3 + 5 + 7 + 6) + (p - 3p - 2p) = 0$$16 - 4p = 0$$4p = 16$$p = 4$.

If $(x + 1)$ is a factor of ${x^4} - (p - 3){x^3} - (3p - 5){x^2} + (2p - 7)x + 6$,then $p = $

Similar Questions

If $x+\frac{1}{x}=2,$ then the value of $x^{2}+\frac{1}{x^{2}}$ will be

When $x+\frac{1}{x}=3$,find $x^{2}+\frac{1}{x^{2}}$.

The condition for the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$ to be equal is

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2}+20=9x$
$II.$ $y^{2}+42=13y$

Let $\alpha$ and $\beta$ be the roots of the quadratic equation $ax^2 + bx + c = 0$. If $a, b, c$ are in $A.P.$ and $\alpha + \beta = 15$,then $\alpha\beta$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module