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

  • Thread starter Angie
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In summary, the conversation was about factoring polynomials and solving a specific problem. The problem given was y^2 - 4y - 5 and the correct answer was determined to be (y-5)(y+1). The conversation also discussed different methods and techniques for factoring polynomials.
  • #1
Angie
4
0
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
 
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  • #2
(y-5)(y+1)
 
  • #3
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.
 
  • #4
FOIL(First Outside Inside Last)-how to remember multiplication. reverse it to factor
 
  • #5
Then the correct answer is:

x^2 - 4y - 4
 
  • #6
no it isn't
 
  • #7
where did the x come from?
 
  • #8
Sorry about that. It is y^2 - 4y -4
 
  • #9
if you are factoring, your answer will more polynomials than you started with. i do not know how you got the y2-4y-4.
 
  • #10
Angie said:
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)
 
  • #11
Thank you for the help guys. :smile:
 

1. What is a factor in algebra?

A factor in algebra is a number or expression that divides evenly into another number or expression, resulting in no remainder. It is used to simplify expressions and solve equations.

2. How do you factor a quadratic equation?

To factor a quadratic equation, you can use the "ac" method or the quadratic formula. The "ac" method involves finding two numbers that multiply to give you the constant term (c) and add to give you the coefficient of the middle term (b). Then, you can use these numbers to rewrite the equation as two separate factors. The quadratic formula, on the other hand, is a formula that can be used to directly find the two factors of a quadratic equation.

3. What is the difference between factoring and simplifying an expression?

Factoring an expression involves breaking it down into its factors, whereas simplifying an expression involves reducing it to its simplest form. Factoring can be seen as the opposite of expanding, where you break an expression into smaller parts, while simplifying focuses on reducing the expression to its most basic form.

4. Can you factor a polynomial with more than one variable?

Yes, polynomials with more than one variable can also be factored. The process is similar to factoring polynomials with one variable, but you will need to consider the terms that are common to both variables and find the greatest common factor (GCF) before factoring.

5. How can factoring be used in real-life situations?

Factoring can be used in various real-life situations, such as calculating the area of a rectangle or finding the prime factors of a number. It is also useful in solving problems in fields such as physics, engineering, and economics. For example, factoring can be used to find the optimal solution in a production problem or to determine the break-even point in a business.

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