(x^4 −x^3 −38x^2 −31x+45)÷(x+5)


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Question Number 58090 by smiak8742 last updated on 17/Apr/19

$(x^{4} - x^{3} - 38 x^{2} - 31 x + 45) \div (x + 5)$
	Answered by Kunal12588 last updated on 17/Apr/19

	$625 + 125 - 950 + 155 + 45 = 0$ $x^{3} - 6 x^{2} - 8 x + 9$ $x + 5) x^{4} - x^{3} - 38 x^{2} - 31 x + 45$ $- x^{4} - 5 x^{3}$ $- 6 x^{3} - 38 x^{2} - 31 x + 45$ $+ 6 x^{3} + 30 x^{2}$ $- 8 x^{2} - 31 x + 45$ $+ 8 x^{2} + 40 x$ $9 x + 45$ $- 9 x - 45$ $0$ $∴ \frac{x^{4} - x^{3} - 38 x^{2} - 31 x + 45}{x + 5} = x^{3} - 6 x^{2} - 8 x + 9$
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