The polynomial p(x) = x^{4} - 2x^{3} + 3x^{2} | vedclass

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

Question

(C) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - c)$,the remainder is $p(c)$.Given $p(x) = x^{4} - 2x^{3} + 3x^{2} -

Vedclass · Accepted Answer

$62$

Vedclass · Answer

(C) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - c)$,the remainder is $p(c)$.Given $p(x) = x^{4} - 2x^{3} + 3x^{2} - ax + 3a - 7$ is divided by $x + 1$,the remainder is $p(-1) = 19$.Substituting $x = -1$ into $p(x)$:$p(-1) = (-1)^{4} - 2(-1)^{3} + 3(-1)^{2} - a(-1) + 3a - 7 = 19$$1 - 2(-1) + 3(1) + a + 3a - 7 = 19$$1 + 2 + 3 + 4a - 7 = 19$$4a - 1 = 19$$4a = 20$$a = 5$Now,we find the remainder when $p(x)$ is divided by $x + 2$,which is $p(-2)$ with $a = 5$:$p(x) = x^{4} - 2x^{3} + 3x^{2} - 5x + 3(5) - 7 = x^{4} - 2x^{3} + 3x^{2} - 5x + 8$$p(-2) = (-2)^{4} - 2(-2)^{3} + 3(-2)^{2} - 5(-2) + 8$$p(-2) = 16 - 2(-8) + 3(4) + 10 + 8$$p(-2) = 16 + 16 + 12 + 10 + 8 = 62$.

The polynomial $p(x) = x^{4} - 2x^{3} + 3x^{2} - ax + 3a - 7$ when divided by $x + 1$ leaves the remainder $19$. Find the value of $a$. Also,find the remainder when $p(x)$ is divided by $x + 2$.

Similar Questions

Evaluate $(153)^{2}$.

Find the zero of the polynomial in the following case:
$p(x) = 3x - 4$

Determine the degree of the following polynomial:
$y^{3}(1-y^{4})$

Factorise the following quadratic polynomial by splitting the middle term:
$x^{2}-12x+20$

Divide $p(x) = x^{3} + 7x^{2} + 14x + 1$ by $x + 3$ and find the quotient and the remainder.

Vedclass Test Series

Exam Paper Generator

Online Exam Module