Question

Which of the following is a factor of 5y^4+4y^3+10y+8?
a. y-5
b.5y-4
c. y^3+2
d. y^2-3
Thanks!!

Answer 1

Do you know about the Rational Root Theorem?

Answer 2

No sir/ma'am I do not.

Answer 3

Well, when you multiply polynomials, the trailing component (\(8\), in this case) and the leading component (\(5y^4\), in this case) are "built up" exclusively from the trailing and leading components of everything you multiplied together.

\[(x+2)(x+3) = x^2+2x+3x+2*3=x^2+5x+6\]6 comes from multiplying 2 and 3
\[(x+1)(x-4)=x^2-4x+1x+(-4)(1) = x^2-3x-4\]-4 comes from multiplying 1 and -4
\[(x+1)(x-4)(x+2) = (x^2-3x-4)(x+2) = \]\[= x^3+2x^2-3x^2-6x-4x-8 = x^3-x^2-10x-8\]-8 comes from multiplying 1, -4, and 2

Answer 4

what this means is that if we are trying to figure out the factors of \(5y^4+4y^3+10y+8\), we can make some intelligent guesses about possible factors by factoring 8 and \(5y^4\).

Answer 5

What are the factors of 8?

Answer 6

The factors of 8 are 1, 2, 4, and 8

Answer 7

right, so that means our choices have to include one of those numbers (or the negative of one of those numbers). For example, there's nothing we can multiply by \((y-5)\) to end up with a polynomial that ends with \(+8\).

Answer 8

So I then start multiplying each option by the factors of 8?

Answer 9

I think i figured out the answer. The answer is 5y-4.

Answer 10

Mmm...don't think so...what do you multiply with \( (5y-4)\) to get \(5y^4+4y^3+10y+8\) ?

Answer 11

The RRT lets us rule out some possible answer choices (a and d) but we still have a few to choose from (b and c).

Answer 12

It would be y^3+2

Answer 13

yes. \[5y(y^3+2) = 5y^4 + 10y\]so we need an additional \[5y^4+4y^3+10y+8-(5y^4+10y+8) = 4y^3+8\]and we can get that by \(4(y^3+2)\) so our factoring would be\[(5y+4)(y^3+2) = 5y^4+4y^3+10y+8\]

Answer 14

If we had tried \(5y-4\) as a factor, we could get a \(y^3\):
\[y^3(5y-4) = 5y^4-4y^3\]leaving us with \[5y^4+4y^3+10y+8-(5y^4-4y^3) = 8y^3+10y+8\]and unfortunately we can't factor out \(5y-4\) from that, so that would be a dead end. Make sense?

Answer 15

Essentially, I used the RRT to eliminate candidates that couldn't possibly work, and then trial division to try the others.

Answer 16

Yes! That makes much more sense now. Thank you very much!! :)

Answer 17

There's another technique called grouping which would also work here:

\[5y^4+4y^3+10y+8 = ( 5y^4 + 10y) + (4y^3 + 8)\]Now we factor each of the groups:
\[5y(y^3+2) + 4(y^3+2)\]Now notice that we've factored the same thing out (but still in parentheses, instead of being out in front) in both terms, so we can rewrite that as \[(5y+4)(y^3+2)\]

Answer 18

Then it's just a matter of looking for the matching answer.