#1**+1 **

1a. \(6x^2+5x-4\)

We can split the middle by having \(6*(-4)=(-24)\). Now, we need to find factors of \(-24\) that add to \(5\). This is \(8\) and \(-3\). So we have \(6x^2-3x+8x-4\). Note that this does not change the value of the equation. Now, we can factor to get \(3x(2x-1)+4(2x-1)\) so we get \((3x+4)(2x-1)\).

1b. \(5x^2-4x-1\).

\(5*(-1)=(-5)\). Factors of \(-5\) include \(-5\) and \(1\). We now can rewrite the equation as \(5x^2-5x+x-1\), which factors into \(5x(x-1)+(x-1)\), so we can get \((x-1)(5x+1)\).

2a. \(x-x^3\)

We can factor out \(x\) first, which gives us \(x(1-x^2)\). Then we can use difference of squares which gives us \(x(1-x)(1+x)\).

2b. \(x^3+2x^2+x\)

We can first factor out \(x\), which gives us \(x(x^2+2x+1)\). Then, we can factor the inside trinomial into \(x(x+1)^2\)

2c. \(x^6+4x^3-5\)

We can split the middle to get \( x^6-3x^3-x^3-5,\) which we can factor into \((x^2+5)(x^3-1)\). Then, we can use difference of cubes on the second term to get \((x^3+5)(x-1)(x^2+x+1)\)

2d. \(16x^4-1\)

We can rewrite this as \((4x)^2-(1)^2\), which we can use difference of squares on to get \((4x^2-1)(4x^2+1)\). Then, we can use difference of squares again to get \((2x-1)(2x+1)(4x^2+1)\)

3a. \(2xy+x+2y+1\)

We can factor out \(x\) from \(2xy+x\). This gives us \(x(2y+1)+(2y+1)\). Then, we get \((x+1)(2y+1)\)

3b. \(x^3+ax^2-x-a\)

We can factor out \(x^2\) from the first two terms. This gives us \(x^2(x+a)-(x+a)\). Then, we get \((x+a)(x^2-1) \)which we can use difference of squares to get \((x+a)(x-1)(x+1)\)

3c. \(x^3+6x^2y+12xy^2+8y^3\)

I'm actually not quite sure about this one, if someone else can do it that would be great. Sorry!

3d. \(x^2+y^2+z^2+2xy-2xz-2yz\)

We can rearrange this and factor some of it to get \((x+y)^2-2z(x+y)+z^2\). Then, we can see that this is a perfect square trinomial and we can factor it to get \((x+y-z)^2\)

-Daisy

dierdurst Mar 16, 2019

#2**+1 **

Thanks, Daisy !!!!

Well done.....

Here's 3(c)

x^3 + 6x^2y + 12xy^2 + 8y^3

(x^3 + 8y^3) + 6x^2y + 12xy^2 factor the first two terms as a sum of cubes

(x + 2y)(x^2 - 2xy + 4y^2) + 6xy ( x + 2y)

(x + 2y ) ( x^2 2xy + 4y^2 + 6xy)

(x + 2y) ( x^2 + 4xy + 4y^2 )

CPhill
Mar 16, 2019