Factor: y^2 - 4y - 5 y^2 - 5y - 4y

[Angie]

I'm looking over today's chapter and I came across this problem I would like to know if I did this correct.


The problem is:

Factor: y^2 - 4y - 5

Answer:

y^2 - 5y - 4y

 

[yourdadonapogostick]

(y-5)(y+1)

 

[Icebreaker]

To factor a polynomial n is to find two other polynomials a and b such that ab = n. It's the same as number factoring.

 

[yourdadonapogostick]

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

 

[Angie]

Then the correct answer is:

x^2 - 4y - 4

 

[yourdadonapogostick]

no it isn't

 

[yourdadonapogostick]

where did the x come from?

 

[Angie]

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

 

[yourdadonapogostick]

if you are factoring, your answer will more polynomials than you started with. i do not know how you got the y²-4y-4.

 

[lurflurf]

I'm looking over today's chapter and I came across this problem I would like to know if I did this correct.


The problem is:

Factor: y^2 - 4y - 5

Answer:

y^2 - 5y - 4y

We want to write y^2 - 4y - 5 in the form
(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)

 

[Angie]

Thank you for the help guys.