Question

find the quotient and remainder when 6x^4+11x^3+x^2-7x+25 is divisible by 3x+4.Also check the remainder obtained by remainder theorem

Answer 1

Hope it will help you well friend

Answer 2

$Let \: p(x) = 6x^{4}+11x^{3}+x^{2}-7x+25 \:and \\g(x) = 3x + 4$

Quotient: 2x³+x²-x-1_

3x+4)6x⁴+11x³+x²-7x+25(

****** 6x⁴ + 8x³

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************ 3x³+x²

************ 3x³ + 4x²

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*************** -3x²-7x

*************** -3x²-4x

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****************** -3x+25

****************** -3x -4

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Remainder = ( 29 )

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Dividend p(x) = 6x⁴+11x³+x²-7x+25

Divisor g(x) = 3x + 4

Quotient q(x) = 2x³+x²-x-1

Remainder r(x) = 29

Verification:

If p(x) is divided by (3x+4 ) then the remainder P(-4/3) .

$Remainder = p(\frac{-4}{3} )\\= 6\Big(\frac{-4}{3}\Big)^{4} + 11\Big(\frac{-4}{3}\Big)^{3} + \Big(\frac{-4}{3}\Big)^{2}-7</p><p>\Big(\frac{-4}{3}\Big)+25 \\= 6\Big(\frac{256}{81}\Big)+11\Big(\frac{-64}{27}\Big)+\Big(\frac{16}{9}\Big)+\frac{28}{3}+25\\= \frac{1536}{81} - \frac{704}{27}+ \frac{16}{9} +\frac{28}{3} + 25 \\= \frac{ 1536 - 2112 +144+756}{81} + 25 \\= \frac{2436-2112}{81} + 25\\= 4+25 \\= 29$

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