basic algebra
posted by John .
1) 2x²7x15
2) 9x²+24x+16
Can you show me all the steps...i don't know how to do it...Thankyou!!:)

first: add 7x to each side. Then divide both sides by 9
second: subtract 9x+16 from each side. Then divide both sides by 15 
we have to factor not solve

1. Y = 2x^27x15 = 0.
Use the AC method of factoring:
A*C = 2*15 = 30 = 1*30 = 3*10.
Choose the pair of factors whose sum = B(7): 3, and 10.
2x^2 + (3x10x)  15 = 0.
Group the 4 terms into 2 factorable pairs:
(2x^210x) + (3x15)
2x(x5) + 3(x5)
(x5) is common to both terms. So we factor out(x5):
(x5)(2x+3) = 0
x5 = 0
X = 5.
2x+3 = 0
2x = 3
X = 1.5.
Solution set: X = 5, and 1.5.
2. 9x^2+24x+16 = 0.
The given expression is a perfect square: 9x^2+24x+16 = (3x+4)^2 = 0.
(3x+4)^2 = 0
Take sgrt of both sides:
+(3x+4) = 0
3x+4 = 0
3x = 4
X = 1 1/3 = 1.33333.
3x4 = 0
3x = 4
X = 1 1/3 = 1.33333.
Solution X = 1.33333, and 1.33333.
Note: The given expression was set to zero, because the solutions are xintercepts and Y = 0 at the xintercepts. 
9x^2+24x+16. Factor only.
9x^2+24x+16 = (3x+4)^2 = (3x+4)(3x+4). 
Thankyou so much!!
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