What must be subtracted from 8x^4 + 14x^3 - 2 | vedclass

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

Question

(C) To find what must be subtracted,we divide the polynomial $P(x) = 8x^4 + 14x^3 - 2x^2 + 8x - 12$ by $D(x) = 4x^2 + 3x - 2$.Step $1$: Divide $8x^4$

Vedclass · Accepted Answer

$15x - 14$

Vedclass · Answer

(C) To find what must be subtracted,we divide the polynomial $P(x) = 8x^4 + 14x^3 - 2x^2 + 8x - 12$ by $D(x) = 4x^2 + 3x - 2$.Step $1$: Divide $8x^4$ by $4x^2$ to get $2x^2$. Multiply $2x^2(4x^2 + 3x - 2) = 8x^4 + 6x^3 - 4x^2$. Subtracting this from $P(x)$ gives $(14x^3 - 6x^3) + (-2x^2 + 4x^2) + 8x - 12 = 8x^3 + 2x^2 + 8x - 12$.Step $2$: Divide $8x^3$ by $4x^2$ to get $2x$. Multiply $2x(4x^2 + 3x - 2) = 8x^3 + 6x^2 - 4x$. Subtracting this from the remainder gives $(2x^2 - 6x^2) + (8x + 4x) - 12 = -4x^2 + 12x - 12$.Step $3$: Divide $-4x^2$ by $4x^2$ to get $-1$. Multiply $-1(4x^2 + 3x - 2) = -4x^2 - 3x + 2$. Subtracting this from the current remainder gives $(12x + 3x) + (-12 - 2) = 15x - 14$.Since the remainder is $15x - 14$,subtracting this from the original polynomial will make it exactly divisible by $4x^2 + 3x - 2$.

What must be subtracted from $8x^4 + 14x^3 - 2x^2 + 8x - 12$ so that the resulting polynomial is exactly divisible by $4x^2 + 3x - 2$?

Similar Questions

If $\alpha, \beta$ and $\gamma$ are the zeros of the cubic polynomial $p(x) = x^{3} + x^{2} - 17x + 15$,then $\alpha\beta + \beta\gamma + \gamma\alpha = \dots$

For the cubic polynomial $p(x) = ax^{3} + bx^{2} + cx + d$; if $a = 3$,$b = -5$,$c = -11$,and $d = -3$,then the cubic polynomial is:

Identify the type of the following polynomial based on its degree: $p(x) = \frac{3}{2} - 0.75x - x^{2}$

$(x+1)$ is not one of the factors of .........

Obtain a quadratic polynomial $p(x) = ax^2 + bx + c$,where the sum of zeros is $\sqrt{2}$ and the product of zeros is $\frac{1}{3}$,given that $a < 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module