How to factor this (probably simple) polynomial?

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willowlinden07

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Hello. I'm reviewing basic algebra I and II concepts for my college algebra class. My original math problem was to factor [imath]10t^4-27t^2-28[/imath]. I converted it to [imath]10u^2-27u-28[/imath] (the "Get Help" section told me to do this). My question is, what would be the best method to factor this? The program I'm using doesn't tell me how to factor it, it only says that I should end up with [imath](5u+4)(2u-7)[/imath]. The internet suggests I use the sum-product pattern, which I've never heard of and don't know how to use. Could someone give me some tips? Thank you.
 
willowlinden07 said:
Hello. I'm reviewing basic algebra I and II concepts for my college algebra class. My original math problem was to factor [imath]10t^4-27t^2-28[/imath]. I converted it to [imath]10u^2-27u-28[/imath] (the "Get Help" section told me to do this). My question is, what would be the best method to factor this? The program I'm using doesn't tell me how to factor it, it only says that I should end up with [imath](5u+4)(2u-7)[/imath]. The internet suggests I use the sum-product pattern, which I've never heard of and don't know how to use. Could someone give me some tips? Thank you.
Click to expand...
Do you recognize that 10u^2 - 27u - 28 is a QUADRATIC function? - If not, do a Google search and tell us what you found?

Have you learned to calculate the "roots" of this equation (that is what is expected here) ?
 
Subhotosh Khan said:
Do you recognize that 10u^2 - 27u - 28 is a QUADRATIC function? - If not, do a Google search and tell us what you found?

Have you learned to calculate the "roots" of this equation (that is what is expected here) ?
Click to expand...
Yep, I know it's quadratic and I know I'm supposed to solve for the roots. I'm mostly wondering if I should factor with the sum-product pattern like Google said (and how to do that) or if there's a better method.
ETA: At the moment I only remember how to use the move-it-to-the-back method or the quadratic formula, but both of those seem overcomplicated for this equation.
 
willowlinden07 said:
Yep, I know it's quadratic and I know I'm supposed to solve for the roots. I'm mostly wondering if I should factor with the sum-product pattern like Google said (and how to do that) or if there's a better method.
ETA: At the moment I only remember how to use the move-it-to-the-back method or the quadratic formula, but both of those seem overcomplicated for this equation.
Click to expand...
Use quadratic formula - that is the most efficient method. What answer/s did you get?
 
willowlinden07 said:
I got 7/3 and -4/5. How would I turn those into (5u+4)(2u−7), which is the answer the program gave me?
I meant 7/2, not 7/3.***
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Your answers about the roots are correct (7/2 & -4/5).

Now think a bit more. How are the roots of quadratic equation and the factored form of the same equation are related?
 
One factored form of your polynomial 10u2−27u−28 is A(u-7/2)(u+4/5) where A =10. You can distribute the 10 appropriately so that you have no fraction. Can you do that? How about factoring the original polynomial?
 
willowlinden07 said:
it only says that I should end up with (5u+4)(2u−7). The internet suggests I use the sum-product pattern, which I've never heard of
Click to expand...
The sum-product method works when the squared term's coefficient is 1. Your polynomial's squared term is 10u^2 not u^2. For such cases, there's an augmentation to the sum-product method called "factor by grouping".

Perhaps you've already used the sum-product method, without knowing a name for it. Given a quadratic polynomial

x^2 + bx + c

that factors nicely, we look for factors of c that combine to make b. In other words, we need two numbers whose sum is b and whose product is c. Maybe that sounds familiar?

Oh, and with the augmented method (factor by grouping), it really helps to understand prime factorizations when working with Integers larger than what you'd see in a multiplication table. For example, with your u-polynomial, factor by grouping requires us to find two factors of -280 that add to make -27.

Subhotosh helped you get the factorization using a formula that you already understood, so that's easiest for you. But, if you're still interested in seeing the other factoring methods, let us know.

PS: Jomo and I would like to see your factorization for the t-polynomial.

?
 
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