# Leaving Cert Math Long Division

Solution to problem

Hi, I'm correcting my work for study, and I cant get my head around this sum.

I understand where the $x^2 + x − cx$ comes from but then when the 6 appears it loses me.

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$(x^2-5x+5cx-6b^2)-(x^2+x-cx)=x(-6+6c)-6b^2$ –  pedja Oct 9 '11 at 8:53
I dont understand shouldn't it be x(−5+5c)−6b^2 –  Rollo Oct 9 '11 at 9:20
Sorry, Sorry, I'm an idiot. –  Rollo Oct 9 '11 at 9:31
Did you try to solve left hand side of the equation that I wrote ? –  pedja Oct 9 '11 at 9:34

$(x^2-5x+5cx-6b^2)-(x^2+x-cx)=x(-6+6c)-6b^2$

because:

$(x^2-5x+5cx-6b^2)-(x^2+x-cx) =$

(by distributing the - sign onto each operand in the bracket this is the equivelant of multilying each term by -1 so -(a+b-c)=-a-b+c)

$(x^2-5x+5cx-6b^2)-x^2 -x + cx =$

re-arranging the terms:

$x^2 - x^2 -x -5x+ cx + 5cx-6b^2 =$

$-6x+ 6cx - 6b^2 =$

$x(-6+6c)-6b^2$

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