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Math Help - Divide 4x^3 + 2x^2 + 10x - 7 by (x-5).

  1. #1
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    Exclamation Divide 4x^3 + 2x^2 + 10x - 7 by (x-5).

    divide 4x^3 + 2x^2 + 10x - 7 by (x-5) using the long division..

    this is polynomial method i think :/.. help please me.. i thought you will understand me.. thanks.. need the answer right now because i will pass this tommorow.. thanks
    Last edited by mr fantastic; October 18th 2010 at 04:21 PM. Reason: Re-titled.
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  2. #2
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    Are you asking to simplify

    \displaystyle{\frac{4x^3 + 2x^2 + 10x - 7}{x-5}}?
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  3. #3
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    A simple way to write "x raised to the 3 power" is x^3.

    In any case, x divides into 4x^3 4x^2 times. Multiplying all of x- 5 by 4x^2 and subtracting, 4x^3+ 2x^2+ 10x- 7- (4x^2- 20x^2)= 22x^2+ 10x- 7.

    x divides into 22x^2 22x times. Multiplying all of x- 5 by 22x and subtracting, 22x^2+ 10x- 7- (22x^2- 110x)= 120x- 7.

    x divides into 120x 120 times. Multiplying all of x- 5 by 120 and subtracting, 120x- 7- (120x- 600)= 593.

    x- 5 divides into 4x^3+ 2x^2+ 10x- 7 gives a quotient of 4x^2+ 22x+ 120 with a remainder of 593.
    Last edited by HallsofIvy; October 20th 2010 at 03:05 AM.
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  4. #4
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    primo candidate for synthetic division ...

    Code:
    5].......4.......2.......10.......-7
    .................20......110.....600
    ---------------------------------------
    .........4.......22......120.....593
    quotient ... 4x^2 + 22x + 120 + \frac{593}{x-5}
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    Or let:

    \displaystyle \frac{4x^3 + 2x^2 + 10x - 7}{x-5} = (ax^2+bx+c)+\frac{r}{(x-5)}

    \displaystyle \Rightarrow 4x^3 + 2x^2 + 10x - 7 = (x-5)(ax^2+bx+c)+r

    \displaystyle \Rightarrow 4x^3 + 2x^2 + 10x - 7 = ax^3+(b-5 a)x^2+(c-5 b)x-5c+r.

    Comparing the coefficients:

    \displaystyle a = 4, b-5a = 2 \Rightarrow b = 22, c-5b = 10 \Rightarrow c = 120

    and finally the remainder -5c+r = -7 \Rightarrow r = 593.

    So \displaystyle \frac{4x^3 + 2x^2 + 10x - 7}{x-5} = (4x^2+22x+120)+\frac{593}{x-5}
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  6. #6
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    Hello, ejaykasai!

    \text{Divide }4x^3 + 2x^2 + 10x - 7\text{ by }(x-5)\text{ using long division.}

    You want Long Division? . . . Here it is . . .


    . . \begin{array}{cccccccccc}<br />
&&&& 4x^2 &+& 22x &+& 120 \\<br />
&& -- & -- & -- & -- & -- & -- & -- \\<br />
x-5 & | & 4x^3 &+& \;\;2x^2 &+& 10x &-& 7 \\<br />
&& 4x^3 &=& 20x^2 \\<br />
&& -- & -- & -- \\<br />
&&&& 22x^2 &+& \;\;10x \\<br />
&&&& 22x^2 &-& 110x \\<br />
&&&& -- & -- & -- \\<br />
&&&&&& 120x &-& \;\;\;7 \\<br />
&&&&&& 120x &-& 600 \\<br />
&&&&&& -- & -- & -- \\<br />
&&&&&&&& 593<br />
\end{array}
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