Factor: 8y^2 +30y +25?

Answer 1

(4y + 5)(2y + 5)

Factor by grouping:

8y^2 + 30y +25

= 8y^2 + 10y + 20y + 25

= 2y(4y + 5) + 5(4y + 5)

= (4y + 5)(2y + 5)

Answer 2

(4y + 5) * (2y + 5)

Just use the quadratic formula to find notional y values:

y = -1¼ and -2½

Now rearrange so it is 0 on the right:

y + 1¼ = 0 and y + 2½ = 0

Now you have the basic factors: (y + 1¼) * (y + 2½) but they not only look aggravating, but do not yet yield the original polynomial. Let's rid ourselves of the fractions by multiplying the first one by 4 on each side and the second by two on each side:

(y + 1¼) * 4 = 0 * 4 and (y + 2½) * 4 = 0 *4
(4y + 5) = 0 and (2y + 5) = 0

Let's put it in factor form again and see if any further adjustments are required to have factors that yield the original polynomial:

(4y + 5) * (2y + 5)

Yep, we're done.

Answer 3

8y^2+30y+25
=8y^2+10y+20y+25
=2y(4y+5)+5(4y+5)
=(4y+5)(2y+5)

Answer 4

(4y+5) (2y+5)

Answer 5

(4y+5) (2y+5)

Answer 6

y = (-30 +/- sqrt(900-800))/16
y = (-30 +/- sqrt(100))/16
y = (-30 +/- 10)/16
y = -40/16 or y = -20/16
y = -2.5 or y = -1.25