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# algebra 2 question?

find m so that x^3+2x^2+mx-3 will leave a remainder of -6 when divided by x+3

### 3 Answers

- Dragon.JadeLv 71 month ago
Hello,

π₯Β³ + 2π₯Β² + ππ₯ β 3

Β Β = π₯Β³ + 3π₯Β² β π₯Β² β 3π₯ + (π + 3)π₯ + 3(π + 3) β 3(π + 3) β 3

Β Β = π₯Β²(π₯ + 3) β π₯(π₯ + 3) + (π + 3)(π₯ + 3) β 3π β 12

Β Β = (π₯ + 3)(π₯Β² β π₯ + π + 3) β 3π β 12

Thus the remainder of the division by (π₯+3) is

Β Β -3π β 12

And if you want it to be equal to -6:

Β Β -3π β 12 = -6

Β Β -3π = 6

Β Β π = -2

Regards,

Dragon.Jade :-)

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- cmcsafeLv 71 month ago
We'll use the remainder theorem.

from x+3 it follows that we must consider the root x-3

Plug it into polynomial and we have the remainder

(-3)^3+2(-3)^2+m(-3)-3 = remainderΒ

We want remainder = -6 soΒ

(-3)^3+2(-3)^2+m(-3)-3 = -6

-27+18-3m-3=-6

m=-2

Check

(x^3+2x^2-2x-3)=(x^2-x+1)*(x+3) -6

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- sepiaLv 71 month ago
(x^3 + 2x^2 + mx + 3) / (x + 3)

=Β x(x - 3)(x + 5)(x + 3)

=Β -6Β

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