Question

find the remainder when the polynomial 2x^4-6x^3+2x^2-x+2 divide by x-2​

Answer 1

Step-by-step explanation:

If x + 2 = 0

x = -2

f(x) = 2x4 – 6x3 + 2x2 – x + 2, [By remainder theorem]

f(x) = 2(-2)4 – 6(-2)3 + 2(- 2)2 – (- 2) + 2

= 2(16) – 6(- 8) + 2(4) + 2 + 2

= 32 + 48 + 8 + 2 + 2 = 92

Hence, required remainder = 92.

plzz mark my answer as the brainliest answer.

Answer 2

Answer:

$\underline{\boxed{\mathsf{Remainder=(-8)}}}$

Step-by-step explanation:

$\underline{\underline{\textsf{Given that}}}$

$\mathsf{Polynomial\: 2 {x}^{4} - 6 {x}^{3} + 2 {x}^{2} - x + 2 \: devided\:by\: x-2 }$

and now we need to find remainder ...

$\underline{\underline{\textsf{Solution}}}$

$\underline{\textsf{Let the polynomial be }}$

$\mathsf{p(x) = 2 {x}^{4} - 6 {x}^{3} + 2 {x}^{2} - x + 2 \:}$

$\bigstar \bf{\textsf{ Remainder theorem}}$

Remainder theorem states that ...

When a polynomial 'p(x)' divided by a linear polynomial 'x - a' then the remainder (r) = p(a)

$\underline{\textsf{Therefore by using remainder theorem in the our question we have}}$

$\mathsf{p(x) = 2 {x}^{4} - 6 {x}^{3} + 2 {x}^{2} - x + 2 \:}$

and

$\mathsf{ Zero\: of\: divisor }$

$\mathsf{ x-2 = 0 \implies x=2}$

Thus

$\mathsf{Remainder\: (r) \:= p(2) = 2 ({2}^{4}) - 6 ({2}^{3}) + 2 ({2}^{2}) - 2 + 2 \:}$

$\implies \mathsf{2 ({2}^{4}) - 6 ({2}^{3}) + 2 ({2}^{2}) - 2 + 2 \:}$

$\implies \mathsf{2 (16) - 6 (8) + 2(4) - 2 + 2 \:}$

$\implies \mathsf{32 - 48 + 8 - 2 + 2 \:}$

$\implies \mathsf{-16+8-2+2= -8}$

$\underline{\textbf{Required remainder is}}$

$\underline{\boxed{\boxed{\textsf{r = (-8)}}}}$