What must be added to the polynomial p(x) = 6 | vedclass

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

Question

(B) To find what must be added to $p(x)$ to make it divisible by $g(x) = 3x^2 - 2x + 4$,we perform polynomial long division of $p(x)$ by $g(x)$.Dividi

Vedclass · Accepted Answer

$17x - 13$

Vedclass · Answer

(B) To find what must be added to $p(x)$ to make it divisible by $g(x) = 3x^2 - 2x + 4$,we perform polynomial long division of $p(x)$ by $g(x)$.Dividing $6x^5 + 5x^4 + 11x^3 - 3x^2 + x + 1$ by $3x^2 - 2x + 4$:$1$. Divide $6x^5$ by $3x^2$ to get $2x^3$. Multiply $2x^3(3x^2 - 2x + 4) = 6x^5 - 4x^4 + 8x^3$. Subtracting this from $p(x)$ gives $9x^4 + 3x^3 - 3x^2 + x + 1$.$2$. Divide $9x^4$ by $3x^2$ to get $3x^2$. Multiply $3x^2(3x^2 - 2x + 4) = 9x^4 - 6x^3 + 12x^2$. Subtracting this gives $9x^3 - 15x^2 + x + 1$.$3$. Divide $9x^3$ by $3x^2$ to get $3x$. Multiply $3x(3x^2 - 2x + 4) = 9x^3 - 6x^2 + 12x$. Subtracting this gives $-9x^2 - 11x + 1$.$4$. Divide $-9x^2$ by $3x^2$ to get $-3$. Multiply $-3(3x^2 - 2x + 4) = -9x^2 + 6x - 12$. Subtracting this gives $-17x + 13$.The remainder is $r(x) = -17x + 13$.To make the polynomial divisible,we must add $-r(x) = -(-17x + 13) = 17x - 13$.

What must be added to the polynomial $p(x) = 6x^5 + 5x^4 + 11x^3 - 3x^2 + x + 1$ so that the resulting polynomial is exactly divisible by $3x^2 - 2x + 4$? (Hint: add $-r(x)$ to $p(x)$)

Similar Questions

Verify that $\frac{-2}{3}$ is a zero of the linear polynomial $p(x)=3x+2$.

Factorize the following polynomial: $p(x) = x^{2} + 5x - 24$

Are the following statements 'True' or 'False'? Justify your answers.
If the graph of a polynomial intersects the $x$-axis at only one point,it cannot be a quadratic polynomial.

One of the factors of $p(x) = x^{3} + 2x^{2} + 3x + 2$ is:

The zero of the linear polynomial $p(x) = ax + b$ is $\ldots \ldots \ldots \ldots$ $(a, b \neq 0)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module