Question

Factoring All Methods

Answer 1

Hello,

Explanation:

`15n^2-110n+200=15(n^2-(22)/(3)x +(40)/(3) )\\`

1. I am thinking about 2 numbers with sum=22/3 and product=40/3.

Those numbers are 4 and 10/3 since 4*10/3=40/3 and 4+10/3=22/3

`15(n^2-(22)/(3)x +(40)/(3) )=15(n-(10)/(3) )(n-4)=5*(3n-10)(n-4)`

2.

`15(n^2-(22)/(3)x +(40)/(3) )=15(n^2-2*(11)/(3) x+((11)/(3))^2 -(121)/(9) +(120)/(9))\\\\=15((n-(11)/(3))^2 -(1)/(9))\\\\=15( n-(11)/(3) -(1)/(3))*(n-(11)/(3) +(1)/(3))\\=15(n-(10)/(3))(n-4)\\\\=5(3n-10)(n-4)\\`

3.

Using discriminant ...

Answer 2

5(3n - 4)(3n - 10)

Explanation:

15n² - 110n + 200 ← factor out 5 from each term

= 5(3n² - 22n + 40) ← factor the quadratic

consider the factors of the product of the coefficient of the n² term and the constant term which sum to give the coefficient of the n- term.

product = 3 × 40 = 120 and sum = - 22

the factors are - 12 and - 10

use these factors to split the n- term

3n² - 12n - 10n + 40 ( factor the first/second and third/fourth terms )

3n(n - 4) - 10(n - 4) ← factor out (n - 4) from each term

(n - 4)(3n - 10)

then

15n² - 110n + 200 = 5(n - 4)(3n - 10)