Leon asked in Science & MathematicsMathematics Β· 1 month ago

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

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  • 1 month ago

    Hello,

    π‘₯Β³ + 2π‘₯Β² + π‘šπ‘₯ βˆ’ 3

    Β Β  = π‘₯Β³ + 3π‘₯Β² βˆ’ π‘₯Β² βˆ’ 3π‘₯ + (π‘š + 3)π‘₯ + 3(π‘š + 3) βˆ’ 3(π‘š + 3) βˆ’ 3

    Β Β  = π‘₯Β²(π‘₯ + 3) βˆ’ π‘₯(π‘₯ + 3) + (π‘š + 3)(π‘₯ + 3) βˆ’ 3π‘š βˆ’ 12

    Β Β  = (π‘₯ + 3)(π‘₯Β² βˆ’ π‘₯ + π‘š + 3) βˆ’ 3π‘š βˆ’ 12

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

    Β Β  -3π‘š βˆ’ 12

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

    Β Β  -3π‘š βˆ’ 12 = -6

    Β Β  -3π‘š = 6

    Β Β  π‘š = -2

    Regards,

    Dragon.Jade :-)

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  • 1 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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  • sepia
    Lv 7
    1 month ago

    (x^3 + 2x^2 + mx + 3) / (x + 3)

    =Β  x(x - 3)(x + 5)(x + 3)

    =Β  -6Β 

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