jjohanar
Answered

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Factoring All Methods

Factoring All Methods class=

Sagot :

Answer:

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

Step-by-step 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)

caylus

Answer:

Hello,

Step-by-step explanation:

[tex]15n^2-110n+200=15(n^2-\dfrac{22}{3}x +\dfrac{40}{3} )\\[/tex]

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

[tex]15(n^2-\dfrac{22}{3}x +\dfrac{40}{3} )=15(n-\dfrac{10}{3} )(n-4)=5*(3n-10)(n-4)[/tex]

2.

[tex]15(n^2-\dfrac{22}{3}x +\dfrac{40}{3} )=15(n^2-2*\dfrac{11}{3} x+(\dfrac{11}{3})^2 -\dfrac{121}{9} +\dfrac{120}{9})\\\\=15((n-\dfrac{11}{3})^2 -\dfrac{1}{9})\\\\=15( n-\dfrac{11}{3} -\dfrac{1}{3})*(n-\dfrac{11}{3} +\dfrac{1}{3})\\=15(n-\dfrac{10}{3})(n-4)\\\\=5(3n-10)(n-4)\\[/tex]

3.

Using discriminant ...

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