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

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(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

$15x - 14$

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(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 $4x^2 + 3x - 2$?

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