dividing polynomials: (x^2 + 8x - 16) / (x + 4)

swampy

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Dec 16, 2008
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the problem is (x^2+8x-16)/(x+4)
the solution i came up with is x-4 but everyone else in my class says there is no solution.
are any of these right at all. see my work belos
x ---- dividing first term by first term to get x
x^2 + 8x - 16
x^2 + 4x ----- multiplying x by divisor x + 4
4x ---- subtracting
4x - 16
4x - 16 --- multiplying 4 by divisor x + 4
0 ---- subtracting

quotient is x + 4
so (x + 4) (x + 4) = x^2 +4x + 4x - 16

is this correct so far???
 
Re: dividing polynomials

swampy said:
the problem is (x^2+8x-16)/(x+4)
the solution i came up with is x-4 but everyone else in my class says there is no solution.
are any of these right at all. see my work belos

x ---- dividing first term by first term to get x
x^2 + 8x - 16
x^2 + 4x ----- multiplying x by divisor x + 4
........4x ---- subtracting
........4x - 16
........4x + 16 --- multiplying 4 by divisor x + 4
...............0 ---- subtracting
...............-32

quotient is x + 4
so (x + 4) (x + 4) = x^2 +4x + 4x - 16<<<< No

(x+4)(x+4) = x^2 +4x + 4x + 16

is this correct so far???<<<< No
 
Re: dividing polynomials

Hello, swampy!

\(\displaystyle (x^2+8x-16) \div (x+4)\)


. . \(\displaystyle \begin{array}{ccccccc}& & & & x & + & 4 \\ & & -- & -- & -- & -- & -- \\ x+4 & ) & x^2 & + & 8x & - & 16 \\ & & x^2 & + & 4x \\ & & -- & -- & -- \\ & & & & 4x & -& 16 \\ & & & & 4x & + & 16 \\ & & & & -- & -- & -- \\ & & & & & - & 32 \end{array}\)


\(\displaystyle \text{Therefore: }\;\frac{x^2+8x - 16}{x+4} \;\;=\;\;x + 4 - \frac{32}{x+4}\)

 
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