Math Hw help plz:factor each expression....

1. 2m square - 11m + 15

#2. 5t sqaure + 28t + 32

1.
2m² - 11m + 15 = 0
2m² - 6m - 5m + 15 = 0
2m(m - 3) - 5(m - 3) = 0
(2m - 5) (m - 3) = 0
2m - 5 = 0 or m - 3 = 0
m = 5/2 or m = 3

2.
5t² + 28t + 32 = 0
5t² + 20t + 8t + 32 = 0
5t(t + 4) + 8(t + 4) = 0
(5t + 8) (t + 4) = 0
5t + 8 = 0 or t + 4 = 0
t = - 8/5 or t = - 4

!!!HOPE I HELPED!!!

1.--- 1.2m^(2)-11m+15

Set up a product of two factors containing two terms each.
( )( )

Since the coefficient of 1.2m^(2) is 1, the coefficients of the left-hand term in each factor must be 1.
(m )(m )

For a polynomial of the form ax^(2)+bx+c, find two factors of c (15) that add up to b (-11).
a=1.2, b=-11, c=15

In this problem -1.67*-7.5=15 and -1.67-7.5=-11, so insert -1.67 as the right hand term of one factor and -7.5 as the right-hand term of the other factor.
(m-1.67)(m-7.5)


2.---- 5t^(2)+28t+32

For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (160) that add up to b (28).
a=5, b=28, c=32

First find the product of a and c to use the ac factoring method.
a*c=160

For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (160) that add up to b (28).In this problem 4*(8)/(5)=(32)/(5) (which is (c)/(a)) and 4+(8)/(5)=(28)/(5) (which is (b)/(a)) , so insert 4 as the right hand term of one factor and (8)/(5) as the right-hand term of the other factor.
(t+4)(t+(8)/(5))

Remove the fraction by multiplying the first term of the factor by the denominator of the second term.
(t+4)(5t+8)

1.
2m² - 11m + 15

2m² - 6m - 5m + 15

2m(m - 3) - 5(m - 3)

(2m - 5)(m - 3)
-------------------------

2.
5t² + 28t + 32
5t² + 20t + 8t + 32

5t(t + 4) + 8(t + 4)

(5t + 8)(t + 4)

1. 2m^2 - 11m + 15 = (2m-5)(m-3)

2. 5t^2 + 28t + 32 = (5t+8)(t+4)

1) (2m - 5)(m - 3)
2) (5t + 8)(t + 4)