How do you evaluate #(5x ^ { 3} + 19x ^ { 2} + 3x - 27) \div ( x + 3)#?

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Answer 1 · 2016-12-09T15:25:21

The quotient is #=5x^2+4x-9# and the remainder #=0#

You can either do a long division or use the remainder theorem

Let's do the long division

#color(white)(aaaa)##5x^3+19x^2+3x-27##color(white)(aaaa)##∣##x+3#

#color(white)(aaaa)##5x^3+15x^2##color(white)(aaaaaaaaaaaaa)##∣##5x^2+4x-9#

#color(white)(aaaaa)##0+4x^2+3x#

#color(white)(aaaaaaa)##+4x^2+12x#

#color(white)(aaaaaaaaaaa)##0-9x-27#

#color(white)(aaaaaaaaaaaaa)##-9x-27#

#color(white)(aaaaaaaaaaaaaaa)##-0-0#

The quotient is #=5x^2+4x-9# and the remainder #=0#

Let's use the remainder theorem

When we divide a polynomial #f(x)# by #x-c#

#f(x)=(x-c)q(x) + r(x)#

When #x=c#

#f(c)=r#

Here,

#f(-3)=5(-3)^3+19(-3)^2+3(-3)-27#

#=-135+171-9-27=0#

The remainder is #=0#