Answer:
- (y +2)(5y +3)
- y = -2, -3/5
Explanation:
You want the factors of 5y² +13y +6, and the solutions to 5y² +13y +6 = 0.
a) Factors
The quadratic ax²+bx+c with a leading coefficient a≠1 can be factored by considering the factors of the product 'ac' that have a sum of 'b'.
5·6 = 30 = (1)(30) = (2)(15) = (3)(10) = (5)(6)
The sums of these factor pairs are, respectively, 31, 17, 13, 11. The factor pair with a sum of 13 is (3)(10).
There are a couple of methods that can be used to apply this knowledge to the factorization of the quadratic. Perhaps the most straightforward is to write the quadratic as 4 terms, using the found values to rewrite the linear term as their sum:
5y² +13y +6 = 5y² +3y +10y +6 . . . . . . . . . 13 ⇒ 3 +10
Now, pairs of terms can be factored:
= y(5y +3) +2(5y +3)
= (y +2)(5y +3) . . . . . . . fully factored expression
b) Solution
The solutions that makes the product of factors be zero will be the values of x that make the factors be zero. A product is only zero if one of the factors is.
y +2 = 0 ⇒ y = -2
5y +3 = 0 ⇒ y = -3/5
The solutions to the quadratic equation are y = {-2, -3/5}.
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Additional comment
Another way to find the factors is this. For factors p, q that have a product of 'ac' and a sum of 'b', the factorization can be ...
(ay +p)(ay +q)/a
For the values in this problem, this would be ...
(5y +3)(5y +10)/5
The latter of these factors can be divided by 5, so this can be simplified to ...
(5y +3)(y +2) . . . . as above