steps 2 factoring?

Question

a. 2n^3-12n^2y+18ny^2

b. 10k^2+105km+270m^2

Answer 1

a) 2n^3-12n^2y+18ny^2

Factor out a 2n

2n(n^2 - 6ny + 9y^2)

Factor the trinomial

2n(n - 3y)(n - 3y)

or

2n(n - 3y)^2

b) 10k^2 + 105km + 270m^2

Factor out a 5

5(2k^2 + 21 km + 54m^2)

Factor the trinomial

5(2k + 9m)(k + 6m)

Answer 2

a) factor out common factors first (always do this)
2n(n^2-6ny+9y^2)
The trinomial is a perfect square (the first and last terms are squares and the middle is twice the product)
so the factorization is 2n(n - 3y)^2

b) Again look for the common factor first
5(2k^2+21km+54m^2)
There are a number of ways to factor this trinomial. One is to multiply the first and last coefficients 2(54) = 108. Find factors of 108 that add to 21. (9 and 12) Rewrite the trianomial as 2k^2+9km+12km+54m^2 and factor by grouping: k(2k+9m) + 6m(2k+9m) = (2k+9m)(k+6m).
So the complete factorization is:
5(2k+9m)(k+6m)

Answer 3

1. look for a common factor, in the first one, its 2n.
2. when this is done, you are left with 2n(n^2-6ny+9y^2)
3. now, you have a perfect square trinomial, meaning that the two factors for the quantity in the parentheses will be the same.
simply separate this into (n-3y)^2 because (-3y * -3y = 9y^2) and (-3ny + -3ny = -6ny)

final answer: 2n(n-3y^2)

same thing for the second one, always look for the greatest common factor first, it will make you life easier during the factoring process because then you will be dealing with smaller numbers.

good luck!

if you still want some assistance, ask your teacher to write out a sample problem step by step, and then have you try and figure out how they did it.

Answer 4

a. 2n(n^2-6ny+9y^2)

2n(n-3y)(n-3y)

b. 10k^2+105km+270m^2

5(2k^2+21km+54m^2)
5(2k+9m)(k+6m)

Answer 5

a)
2n(n² - 6ny + 9y²)
2n(n - 3y)²

b)
5(2k² + 21km + 54m²)
5(2k + 9m)(k + 6m)

Answer 6

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