Question

4x^4-5x^3-7x+1÷(4x-1)

Answer 1

HEY Buddy....!! here is ur answer

(4x⁴–5x³–7x+1) ÷ (4x–1)

=> $\frac{4 {x}^{4} - 5 {x}^{3} - 7x + 1 }{4x - 1} \\ \\ = > {x}^{3} - {x}^{2} - \frac{1}{4} x - \frac{29}{16} = Quotient \: and \: (- \frac{13}{16} )= Remainder$

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Answer 2

Answer:

Quotient = $4x^3 -x^2- \frac{x}{4}-\frac{29}{16}$

Remainder = $-\frac{13x}{16}$

Step-by-step explanation:

Given expression,

$(4x^4 - 5x^3 - 7x + 1)\div (4x-1)$

By long division method ( shown below )

We get,

$\frac{4x^4 - 5x^3 - 7x + 1}{4x-1}=4x^3 + \frac{x^2}{4}+\frac{x}{16}+\frac{113x/16 + 1}{4x-1}$

That is,

Quotient = $4x^3 -x^2- \frac{x}{4}-\frac{29}{16}$

Remainder = $-\frac{13x}{16}$

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