- #1

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The problem is:

Factor: y^2 - 4y - 5

Answer:

y^2 - 5y - 4y

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- Thread starter Angie
- Start date

- #1

- 4

- 0

The problem is:

Factor: y^2 - 4y - 5

Answer:

y^2 - 5y - 4y

- #2

- 258

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(y-5)(y+1)

- #3

Icebreaker

- #4

- 258

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FOIL(First Outside Inside Last)-how to remember multiplication. reverse it to factor

- #5

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Then the correct answer is:

x^2 - 4y - 4

x^2 - 4y - 4

- #6

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no it isn't

- #7

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where did the x come from?

- #8

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Sorry about that. It is y^2 - 4y -4

- #9

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- #10

lurflurf

Homework Helper

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We want to write y^2 - 4y - 5 in the formAngie said:

The problem is:

Factor: y^2 - 4y - 5

Answer:

y^2 - 5y - 4y

(a y+b)(c y+d)=a c y^2+(a d+b c)y+b d

hence find numbers a,b,c,d such that

a c=1

a d+b c=-4

b d=-5

First we determine the prime factos of 1 and 5 as the middle term 4 is harder to deal with.

1 has no prime factors the only way 1 can be written as a product of natural numbers is 1*1 so a=c=1

5=1*5 so we chose d and b to be either b=-1,d=5 or b=1,d=-5 we guess one and if we are wrong it was the other one.

lets guess b=-1,d=5

a d+b c=1*5+(-1)*1=5-1=4 so we guessed wrong

let b=1,d=-5

a d-b c=1*(-5)+1*1=-5+1=-4

right! so

y^2-4y-5=(y+1)(y-5)

another way to see this is

write 4 as 5-1 because 5 is a factor of 5 and 1 is a factor of 1

y^2-4y-5=y^2-(5-1)y-5

=y^2+1-5y-5

=(y^2+1)+(-5y-5)

=y(y+1)-5(y+1)

=(y-5)(y+1)

- #11

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Thank you for the help guys.

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