Pre-Alg help with factoring?

Question

factor by first the greatest common factor, then factor the trinomial that remains.?

2x^3-14x^2+20x

factor the greatest common factor

3a^4-18a^3b+27a^2b^2

72y^2+60y-72

factor by the greatest common factor for the trinomial.

2x^2(x+2)+13x(x+2)+15(x+2)

Answer 1

factor by first the greatest common factor, then factor the trinomial that remains.?

2x^3 - 14x^2 + 20x
Each part of this is divisable by 2x (the greatest common factor) so pull that out (divide each by 2x), and you have left
x^ - 7x +10
now factor that:
(x - 5)(x - 2)

factor the greatest common factor = find the greatest common factor of the coefficient and the first variable and the second variable in each term

3a^4-18a^3b+27a^2b^2
coeffieient GCF = GCF of 3, 18, 27 = 3
the lowest exponent of a in the terms = a^2
since there is no b in the first term we won't factor out any b's
so the greatest common factor is 3a^2
divide each term by that and you have left
a^2 - 6ab + 9b^2

72y^2+60y-72
coeffieient GCF = GCF of 72, 60, 72 = 12
since there is no y in the last term we won't factor out any y's
so the greatest common factor is 12
divide each term by that and you have left
6y^2+5y-6

factor by the greatest common factor for the trinomial.

2x^2(x+2)+13x(x+2)+15(x+2)
immediatly you should see that there is (x+2) in each term so we will put that in GCF
we now have 2x^2 + 13x + 15
there is nothing else we can take out so we are done
GCF = (x+2)
2x^2 + 13x + 15

BTW
Ang is a brag, and no one likes that

Answer 2

For 2x^3-14x^2+20x take 2x out as a common factor which gives 2x(x^2-7x+10). Factorising the contents of the bracket by finding factors of 10 which sum to -7 (-5x,-2) Gives the final solution of 2x(x-2)(x-5)
For 3a^4-18a^3b+27a^2b^2 take 3a^2 out as a common factor leaving 3a^2(a^2-6ab+9b^2). The contents of the bracket is a perfect square ie a^2-6ab+(3b)^2 which factorises to (a-3b)^2. The final solution is 3a^2(a-3b)^2
For 72y^2+60y-72 take 12 as a common factor leaving12(6y^2+5y-6) Factorise the bracket by multiplying the 6 and -6 to get -36. Now the factors of-36 which sum to 5 are -9 and 4. Take the expression in the bracket and split the middle term into 2 parts using the -9 and 4 ie 6y^2+5y-6=6y^2-9y+4y-6. The value of the expression has not changed as -9y+4y=5y. Now factorise the new expresision by pairing the terms. (6y^2-9y)+(4y-6)=3y(2y-3) +2(2y-3). Now the bracket is a common factor so taking this out the front gives (2y-3)(3y+2).
For the last expression take the bracket out as a common factor giving (x+2)(2x^2+13x+15) Now 2x15=30 the factors of 30 which sum to 13 are 10 and 3 so the expressioncan be turned into (2x^2+10x)+(3x+15). Factorising in pairs gives 2x(x+5)+3(x+5). The bracket is now a common factor so taking it out gives (x+5)(2x+3). Hope this helps.

Answer 3

ok...I'm not going to give you the answers because if this is homework, I will not aid you in cheating...BUT...I will tell you how to do them.

The first one...you are basically trying to simplify the problem at first. You need to use the distributive property (GCF) and pull it to the outside of the problem. Think about it like this. The coefficients (the number in front of a variable) are all even. I suggest you try 2. Then, you see that all of them have an X with them. So maybe try pulling out 2x and go from there.

As for factoring a trinomial. It took me awhile for this to click for me. What I do is come up with the factors for the last number and the factors for the first number(and variable) and see which combinations make the middle number(and variable)

This is really hard to explain without seeing you work on it so if you want to IM me you can and I can help you step by step.

(I am in college by the way and Math is one of my best subjects...I got 100+ in both my college math courses and graduated 7th in my class in high school!)

~Ang

BTW...I wasnt bragging...I wanted to make sure you knew I had credentials before you just believed me. Anyone can type numbers and letters. I wanted you to be sure that I wasn't just being a know it all.

Oh and I think it is wrong for people to just give you the answers. Like I said, if you need help with this stuff in the future, I would love to help you!

Answer 4

the first situation is of the variety ac + bc the position c = (2x - 3y), so that you'll element it out ac + bc = c(a + b) 10x^3(2x - 3y) - 15x^2(2x - 3y) = (2x - 3y)(10x^3 - 15x^2) If we interchange the order (10x^3 - 15x^2)(2x - 3y) we may be able to element out a 5x^2 5x^2(2x - 3)(2x - 3y) ---------------- 5xy^2 + 5y^2 + 3ax + 3a element out 5y^2 from the first 2 words, and 3a from the terrific 2 words 5y^2(x + a million) + 3a(x + a million) it really is like the ac + bc --------------------- 3x^3 + 2x^2 - 27x - 18 it is a touch trickier through - signs and indicators element out x^2 from the first 2 words, and -9 from the terrific 2 words x^2(3x + 2) - 27(3x + 2) appears like ac + bc back be conscious the signal substitute - if we take -27(3x + 2) and do the multiplication, we get -27x - 18, it really is what we began with --------------------- y^2 - y - 6 back, that is functional to look at a progression making use of better favourite expressions as an social gathering, (x + a)(x + b) = x^2 + ax + bx + ab or x^2 + (a + b)x + ab For both constants a & b, the coefficient of the x time period is their sum and the consistent time period in the trinomial is their product y^2 -y -6 the elements of - 6 are 2, -3 or -2, 3 The sum of two and -3 provides us the -a million coefficient of the y time period y^2 - y - 6 = (y + 2)(y - 3)

Answer 5

2x(x-5)(x-2)

there's a start