Factoring Help!!! Pre- Alg?

Question

factor the greatest common factor

10x^3(2x-3y)-15x^2(2x-3y)

factor by grouping...?

5xy^2+5y^2+3ax+3a

3x^3+2x^2-27X-18

Factor trinomial

y^2-y-6

Answer 1

1) (2x^2 - 3x) (5x(2x-3y)

2) (5y^2 + 3a) (x + 1) .... edited as I misread it as x^2 y^2

3) (x-3)^2 (x - 2/3)

4) (y-2)(y+3)

Answer 2

The first problem is of the form

ac + bc

where c = (2x - 3y), so you can factor 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 can factor out a 5x^2

5x^2(2x - 3)(2x - 3y)

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5xy^2 + 5y^2 + 3ax + 3a

Factor out 5y^2 from the 1st 2 terms, and 3a from the last 2 terms

5y^2(x + 1) + 3a(x + 1)

This is like the ac + bc

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3x^3 + 2x^2 - 27x - 18

This is a little trickier because of the - signs

Factor out x^2 from the 1st 2 terms, and -9 from the last 2 terms

x^2(3x + 2) - 27(3x + 2)

Looks like ac + bc again

Note the sign change - if we take -27(3x + 2) and do the multiplication, we get -27x - 18, which is what we started with

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y^2 - y - 6

Again, it is useful to look at a pattern using more general expressions

For example,

(x + a)(x + b) = x^2 + ax + bx + ab

or x^2 + (a + b)x + ab

For the 2 constants a & b, the coefficient of the x term is their sum and the constant term in the trinomial is their product

y^2 -y -6

The factors of - 6 are 2, -3 or -2, 3

The sum of 2 and -3 gives us the -1 coefficient of the y term

y^2 - y - 6 = (y + 2)(y - 3)

Answer 3

Given Equation

= 10x ^ 4-15x^ 2 (2x- 3y)
Taking the common factor (2x - 3y) out

We get (2x - 3y) (10x^ 3 - 15 x^ 2)

= 5x^ 2 { 2x - 3) (2x - 3y)}

So the factors of the Equation are

5x ^2, (2x - 3), (2x-3y)

Answer 4

factor the greatest common factor

10x^3(2x - 3y) - 15x^2(2x - 3y)
(10x^3 - 15x^2)(2x - 3y)
(5x^2)(2x - 3)(2x - 3y)

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factor by grouping...?

5xy^2 + 5y^2 + 3ax + 3a
(5xy^2 + 5y^2) + (3ax + 3a)
5y^2(x + 1) + 3a(x + 1)
(5y^2 + 3a)(x + 1)

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3x^3 + 2x^2 - 27x - 18
(3x^3 + 2x^2) + (-27x - 18)
x^2(3x + 2) - 9(3x + 2)
(x^2 - 9)(3x + 2)
(x - 3)(x + 3)(3x + 2)

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Factor trinomial

y^2 - y - 6 = (y - 3)(y + 2)

Answer 5

For 2x^3-14x^2+20x take 2x out as a common ingredient which provides 2x(x^2-7x+10). Factorising the contents of the bracket via finding factors of 10 which sum to -7 (-5x,-2) provides the proper answer of 2x(x-2)(x-5) For 3a^4-18a^3b+27a^2b^2 take 3a^2 out as a common ingredient leaving 3a^2(a^2-6ab+9b^2). The contents of the bracket is a proper sq. ie a^2-6ab+(3b)^2 which factorises to (a-3b)^2. the proper answer is 3a^2(a-3b)^2 For 72y^2+60y-seventy two take 12 as a common ingredient leaving12(6y^2+5y-6) Factorise the bracket via multiplying the 6 and -6 to get -36. Now the factors of-36 which sum to 5 are -9 and four. Take the expression interior the bracket and split the midsection term into 2 areas using the -9 and four ie 6y^2+5y-6=6y^2-9y+4y-6. the cost of the expression has no longer replaced as -9y+4y=5y. Now factorise the recent expresision via pairing the words. (6y^2-9y)+(4y-6)=3y(2y-3) +2(2y-3). Now the bracket is a common ingredient so taking this out the front provides (2y-3)(3y+2). For the final expression take the bracket out as a common ingredient 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 became into (2x^2+10x)+(3x+15). Factorising in pairs provides 2x(x+5)+3(x+5). The bracket is now a common ingredient so taking it out provides (x+5)(2x+3). wish this facilitates.

Answer 6

5x^2((2x-3)^2)

(5y^2+3a)(x+1)

x(3x^2+2x-27)-18

(y-3)(y+2)