36y^{4}(y + 12)^{3} + y^{5}(y + 12)^{4}

36y^{4}(y + 12)^{3} + y^{5}(y + 12)^{4}

Let (y + 12) = u

Now rewrite the expression:

36y^{4}u^{3 + }y^{5}u^{4} GCF: y^{4}u^{3
}

y^{4}u^{3}(36 + yu)

Now, for the sake of simplicity, revert u=(y+12) inside the parenthesis only:

y^{4}u^{3}(36 + y(y + 12))

Now, expand the expression inside the parenthesis:

y^{4}u^{3}(36 + y^{2} + 12y)

Now, rearrange the expression inside the parenthesis:

y^{4}u^{3}(y^{2} + 12y + 36)

Look at that expression inside the parenthesis... is it a perfect square? Indeed it is!! 36 = 6^{2 }and 12y = 2(6)y. Then, factor it:

y^{4}u^{3}(y + 6)^{2}

Finally... do you remember u = (y+12)? Now it's the time to revert the other u:

y^{4}(y + 12)^{3}(y + 6)^{2} Too much fun, isn't it?

************************************

x^{2} - 10x + 24

This trinomial may be factored as the product of two binomials if we find two numbers such that:

- Their product is equal to 24, and

- Their sum is equal to -10

These two numbers indeed exist:

they are -6 and -4: (-6)(-4) = 24; and (-6)+(-4) = -10

Then our trinomial factors neatly: (x - 6)(x - 4)

## Comments

#1: Follow the same pattern: 2 numbers such that their product is -30 and their sum is 1. They're -6 and 5. Then: (x-6)(x+5)

#2: (9x+4)(3x+8)

#3: Difference of 2 squares: (6t+5s)(6t-5s)

#4: Factor by CF: 6x(x

^{2 - }16) Now difference of 2 squares: 6x(x-4)(x+4)#5: Same as #4: x

^{6}y^{5}(x^{2 }- y^{2}); then x^{6}y^{5}(x - y)(x + y)