Factoring Polynomials take 2

[BlBl]

I was told to start a new thread so here it is:

Ok, here's a few more (all factoring):

\(\displaystyle 3(5x+2)^2(5)(3x-4)^4+(5x+2)^3(4)(3x-4)^3(3)\)

\(\displaystyle 5(x^2+4)^4(8x-1)^2(2x)+2(x^2-4)^5(8x-1)(8)\)

I assumed that the GCF for the first was \(\displaystyle (5x+2)^2(4)(3x-4)^3\). With that I… was stuck. I didn't know what to do with those weird numbers off by themselves in parentheses for this one and the next one. Nothing in the book covers that specifically. All I need to know is a rule. I already have the answer.

Help?

These might be more tedious than difficult.

 

[soroban]

Hello, BlBl!

\(\displaystyle 3(5x+2)^2(5)(3x-4)^4 + (5x+2)^3(4)(3x-4)^3(3)\)

Simplify: .\(\displaystyle 15(5x+2)^2(3x-4)^4 + 12(5x+2)^3(3x-4)^3\)


We have two groups: .\(\displaystyle \underbrace{15(5x+2)^2(3x-4)^4} + \underbrace{12(5x+2)^3(3x-4)^3}\)

They have common factors: .\(\displaystyle 3,\;(5x+2)^2,\text{ and }\,(3x-4)^3\)
Factor them out!

. . . . \(\displaystyle 3(5x+2)^2(3x-4)^3\,\big[5(3x-4) + 4(5x+2)\big] \)

. . \(\displaystyle =\;3(5x+2)^2(3x-4)^3\,\big[15x - 20 + 20x + 8\big] \)

. . \(\displaystyle =\;3(5x+2)^2(3x-4)^3\,\big(35x - 12\big)\)

\(\displaystyle 5(x^2+4)^4(8x-1)^2(2x) + 2(x^2-4)^5(8x-1)(8)\)
. . . . . . . . . . . . . . . . . . . . . . . . . . .?

Is there a typo in this one?
I assume both quadratic factors are \(\displaystyle (x^2+4).\)

Simplify: .\(\displaystyle \underbrace{10x(x^2+4)^4(8x-1)^2} + \underbrace{16(x^2+4)^5(8x-1)}\)

The two groups have common factors: .\(\displaystyle 2,\;(x^2+4)^4,\text{ and }(8x-1)\)
Factor them out!

. . . . \(\displaystyle 2(x^2+4)^4(8x-1)\,\big[5x(8x-1) + 8(x^2+4)\big] \)

. . \(\displaystyle =\; 2(x^2+4)^4(8x-1)\,\big[40x^2 - 5x + 8x^2 + 32\big]\)

. . \(\displaystyle = \;2(x^2+4)^4(8x-1)\,\big(48x^2 - 5x + 32 \big)\)

 

[jsterkel]

Simplify your life: let u = 5x+2, v = 3x - 4
So: 15 u^2 v^4 + 12 u^3 v^3
Kapish? Continue...

Nice!

 

[BlBl]

Thanks you all ^_^ and yes, Soroban, there was a typo, and you were correct in your correction.

So it seems the rule that I'm missing is to take the single numbers and multiply them together then put them before the expressions. I suppose this works because in multiplication the arrangement of the problem is irrelevant.

Ok, then I'm good on my polynomials chapter. I can do everything they ask me to do now, without hesitation. And so long as I'm careful and take my time I should be perfect.

Thanks again!