# Simplify y^2 + 8y + 25?

I cant figure out two numbers that add up to 8 and multiply to 25.

Help.

Update:

Sorry! I meant factor.

Relevance
• Dr Pi
Lv 4

y² + 8y + 25 Plug into quadratic formula: (-b±√(b²-4ac))/(2a) where ax² + bx + c = 0

= (-8±√((-8)²-(4)(1)(25))/(2(1))

= (-8±√(-36))/(2)

= (-8±6i)/(2)

= -4±3i

= -4+3i & -4-3i

Change to factors by subtract these #'s from y

(y-(-4+3i))(y-(-4-3i))

= (y+4-3i)(y+4+3i)

To check answer plug in a # & see if they both come out to be the same. I'll plug in 5 for y.

(5)^2 + 8(5) + 25

= 25 + 40 + 25

= 90

(5+4-3i)(5+4+3i)

= (9-3i)(9+3i)

= 81 + 27i - 27i - 3i²

= 81 + 9

= 90

There are no two numbers that add up to 8 and multiply to 25

So you must complete the square

(y² + 8y + 16) + 25 - 16

(y + 4)² + 9

There are no two numbers that add up to 8 and multiply to 25 so you must

Complete the square

(y² + 8y + 16) + 25 - 16

(y + 4)² + 9

• JOS J
Lv 7

25 + y (8 + y)

• Anonymous

y^2 + 8y + 25 = 0

this is a second degree equation so it must have 2 solutions

solution number 1:

y = - 8 + sqrt ( 64 - 100) / 2

y = - 8 + sqrt (-36) / 2

y = -8 + 6i / 2

y = - 4 + 3i

where i = sqrt (-1)

now lets pretend that solution n. 1 (-4 + 3i) equals = a

solution number 2:

y = - 8 - sqrt ( 64 - 100) / 2

y = - 8 - sqrt (-36) / 2

y = -8 - 6i / 2

y = - 4 - 3i

now lets pretend that solution n. 2 (-4 - 3i) equals = b

---

when you have

(y-a)(y-b)

y equals = + a and + b

so let's exchange a and b with the solutions of the previous equation

( y - ( - 4 + 3i ) ) ( y - ( - 4 - 3i ) ) = 0

(y+4-3i)(y+4+3i)=0 <----------------------- THIS IS THE ANSWER