[SOLVED] Help factor 4((m+n)^3)-4 and 36x^2-9y^2+6y-1

**redhawks** · February 12th 2009, 04:26 PM

only 2 i cant get

**skeeter** · February 12th 2009, 04:35 PM

$4(m+n)^3 - 4 =$

$4[(m+n)^3 - 1] =$

note the difference of cubes in the brackets ...

$4[(m+n)^3 - 1^3]$

remember how to factor the difference of two cubes?

$36x^2-9y^2+6y-1=36x^2 - (9y^2 - 6y + 1)$

$36x^2 - (3y - 1)^2$

difference of two squares, correct?

**Soroban** · February 12th 2009, 04:38 PM

Hello, redhawks!

$4(m+n)^3 - 4$

We have: . $4\underbrace{\bigg[(m+n)^3 - 1\bigg] }_{\text{diff. of cubes}}$

. . . $= \;\bigg[(m+n)-1\bigg]\,\bigg[(m+n)^2 + (m+n) + 1\bigg]$

. . . $= \;(m+n-1)(m^2 + 2mn + n^2 + m + n + 1)$

$36x^2 - 9y^2 + 6y - 1$

We have: . $36x^2 - (9y^2 - 6y + 1)$

. . . $= \;36x^2 - (9y^2 - 6y + 1)$

. . . $= \;\underbrace{36x^2 - (3y - 1)^2}_{\text{diff. of squares}}$

. . . $= \;\bigg[6x - (3y-1)\bigg]\,\bigg[6x + (3y - 1)\bigg]$

. . $= \;(6x - 3y + 1)(6x + 3y - 1)$

Edit: Too slow . . . again!
.

**mr fantastic** · February 12th 2009, 04:39 PM

Originally Posted by redhawks

only two i cant get

$4 \left[ (m+n)^3 - 1\right]$ and factorise the difference of two cubes.

$36x^2 - (9y^2 - 6y + 1) = (6x)^2 - (3y - 1)^2$ and factorise the difference of two perfect squares.

**redhawks** · February 12th 2009, 04:40 PM

Thanks to both of you

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[SOLVED] Help factor 4((m+n)^3)-4 and 36x^2-9y^2+6y-1

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