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Goshia [24]
2 years ago
7

The factor of 16y² + 62y - 45 = 0?

Mathematics
2 answers:
Alex17521 [72]2 years ago
5 0

Step-by-step explanation:

we are looking for an expression like

(ay + b)(cy + d)

so that when doing all the multiplications this is equal to our original expression.

that is called factoring, as we are finding the 2 terms that when multiplied with each other create the original expression. we are finding the factors.

and we are normally looking for integer values for a, b, c, d.

ay×cy + b×cy + d×ay + b×d

acy² + (bc + da)y + b×d = 16y² + 62y - 45

so, we know that one of b, d must be negative, the other positive.

and a, c must have the same sign, so that their product is positive.

anyway, we compare now

ac = 16

bc + da = 62

bd = -45

the possible factors of 45 are (remember the different signs)

1 45

3 15

5 9

the possible factors of 16 are (remember they need the same signs)

1 16

2 8

4 4

let's start with the assumption a=4, c=4

is there a fitting factor pair of 45 to allow us to get

bc + da = 62 ?

b×4 + d×4 = 62

b + d = 62/4 = 15.5

since b and d should be integer, we cannot get a non-integer as their sum. so, that does not work.

assume a = 2, c = 8

b×8 + d×2 = 62

4b + d = 31

and here we find with a little trial and error with the factor pairs of 45 that

b = 9 and d = -5 fit that criteria.

4×9 - 5 = 36 - 5 = 31

perfect.

so, our factoring is

(2y + 9)(8y - 5)

control :

2y×8y + 2×-5y + 9×8y + 9×-5 =

= 16y² -10y + 72y - 45 = 16y² + 62y - 45

correct.

Alla [95]2 years ago
4 0
Ans= (8y-5)(2y+9)=0
Explanation-
16y^2+62y-45=0

16y^2+72y-10y-45=0
8y(2y+9) -5(2y+9)=0
(8y-5)(2y+9)=0
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