How to factor trinomial 9y^4-54y^3+45y^2?

Answer 1

9y²(y² - 6y + 5)
9y²(y-5)(y-1)

Answer 2

First, pull out the common factor of 9y^2. This gives you,

9y^2(y^2-6y+5). Then, split up this up using the normal method for factoring a second-degree trinomial.

9y^2(y-1)(y-5)

Answer 3

9y^4 - 54y^3 +45y^2 factor out 9y^2
9y^2(y^2-6y+5)
9y^2(y-5)(y-1)

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Answer 4

Factor out a 9y^2:

9y^2(y^2 - 6y + 5)

Now factor y^2 - 6y + 5, which factors nicely. The factorization has the form (y+a)(y+b), where ab=5, and a+b=-6. Both numbers must have the same sign, since their product is positive, and since their sum is negative, they must be negative.

Answer 5

9y^4 - 54y^3 + 45y^2

factor out 9y^2

9y^2 (y^2 - 6y + 9)
(y-3)^2 = (y-3)(y-3) = y^2 - 2(3y) + 3^2
= 9y^2 (y-3)^2

Answer 6

9y^4 - 54y^3 + 45y^2
9y^2(y^2 - 6y + 5)
(9y^2)(y - 5)(y - 1)

Answer 7

factor out a 9

9(y^4 - 6y^3 +5y^2)

factor out a y^2

9y^2(y^2-6y+5)

the "-6" coefficient in front of the "y" makes it obvious that you need a "-5" and a "-1" thus

(9y^2)(y-5)(y-1)

you should always check your work by multipling back out the (y-5)(y-1) to prove that it does equal (y^2-6y+5)

Answer 8

9y^4-54y^3+45y^2 dividy by 9y^2
9y^2(y^2-6y+5)
9y^2(y-5)(y-1)

Answer 9

synthetic division....

1 | 9 -54 45 0 0
| 9 -45 0 0
|_____________
9 -45 0 0 0

(9y^3-45y^2)(y-1)

5| 9 -45 0 0
| 45 0 0
|___________
9 0 0 0

*****9y^2(y-5)(y-1)*****


or.....

divide all of the numbers by 9y^2

9y^2(y^2-6y+5) then reverse FOIL...

*****9y^2(y-5)(y-1)*****

Answer 10

ok first, you can take out 9y^2 from each part:

9y^2(y^2-6y+5)

then you can factor as you normally would:

9y^2 (y-1)(y-6)

I hope that helped!

Answer 11

i don't have self belief it is factored. If the 40 two became useful, it ought to be factored. till somebody can discover something I neglected, i won't hit upon a factors of -40 two that as quickly as expanded by using 2 upload as much as -26.