How to factor a polynomial with 4 terms

[juanista19]

I have this polynomial:
How do I factor it? All I know is how to factor a trinomial by either using PSA 1 or 2, but I'm having a hard time doing this one. Can someone help please? Thank you.

 

[tkhunny]

#1 - You try something, anything. Thoughts?
#2 - General principle, if f(a) = 0, then (x-a) is a factor. Think about that f(a) = 0 really hard.
#3 - A quick look at signs and exponents can help. If all the exponents were even, and all the signs positive, please don't spend ANY time looking for real roots or real-valued factors.
#4 - If there are RATIONAL-valued factors, they can be only +/-1, +/- 1/2 +/- 3, or +/- 3/2 -- You tell me why.
#5 - Being cubic, once you find one factor, the other two should be easy. It's your chance to demonstrate PSA 1 or 2.
#6 - If you really can't track down a factor, there are other ways to go about it.

Give it a go!

 

[Steven G]

Try x=1 and x= -1. Why is it obvious x=1 will not work?

 

[Harry_the_cat]

What is PSA 1 or 2?

 

[ISTER_REG]

As previous speakers have noted. A trivial idea is to insert x-values, like [MATH]x=-3[/MATH] to [MATH]+3[/MATH] in one step and check if the zero value appears. This is sufficient for school and for university in most cases.

If you insert the values [MATH]x= -3[/MATH] or [MATH]x= -1[/MATH], then you get the following for [MATH]x=-3[/MATH]:
[MATH]2(-3)^3 + 9(-3)^2 +10(-3) + 3 = 0[/MATH]. This is a true statement, so one solution is [MATH]x = -3[/MATH]. The same is then true for [MATH]x = -1[/MATH]. If you have these two factors, you can split them off from the polynomial as linear factors.

For [MATH]x= -3[/MATH] you can split off the following linear factor [MATH](x- (-3)) = (x+3)[/MATH]. Now perform a polynomial division on the polynomial with the linear factor [MATH](x+3)[/MATH]. So [MATH](2x^3+9x^2+10x+3) \div (x+3) = 2x^2+3x+1[/MATH]
Thus the original polynomial has been reduced considerably. Remember that [MATH](2x^2+3x+1)(x+3) = 2x^3+9x^2+10x+3[/MATH].

Important: My explanations here are useful for many polynomials, but there are also exceptions, where this does not work out as well as in this case.

Spoiler: The full factorization of your polynomial is
[MATH]2x^3+9x^2+10x+3 = (2x+1)(x+1)(x+3)[/MATH]

 

[tkhunny]

Important: My explanations here are useful for many polynomials, but there are also exceptions, where this does not work out as well as in this case.

What exceptions? This case works just fine simply finding one factor at a time and reducing the polynomial. Once you get it down to Quadratic, you are done.

It's not a MONIC polynomial. Is that why you think it's an exception?

Given:
f(x) = 2x^3+9x^2+10x+3
Try x = -3. This is listed among ALL the possibilities for RATIONAL roots way back in my #4.
Indeed, we get f(-3) = 0, thus x+3 is a factor.
f(x) = (x+3)(2x^2 + 3x + 1)
That second factor is Quadratic, factor using your favorite method:
2x^2 + 3x + 1 = (2x + 1)(x + 1)
f(x) = (x+3)(2x + 1)(x + 1)
I'm not seeing the exception.

If there are NO Rational Roots, it's a different problem. If there are rational roots, just find them. In a textbook problem, there aren't usually that many to try.

Please use Synthetic Division (aka Horner's Method) to evaluation the polynomial.
f(-3) = ((2(-3) + 9)*(-3) + 10)*(-3) + 3
f(-3) = ((-6 + 9)*(-3) + 10)*(-3) + 3
f(-3) = ((3)*(-3) + 10)*(-3) + 3
f(-3) = (-9 + 10)*(-3) + 3
f(-3) = (1)*(-3) + 3
f(-3) = -3 + 3
f(-3) = 0

 

[ISTER_REG]

@tkhunny I tried to point out that certain polynomials cannot be factorized so "nicely". This guessing the solutions about, for example, a table of values for different x works here quiet well. But if the solutions are not so obvious, then one must fall back on numerical procedures. That is what I wanted to imply with it

 

[tkhunny]

My #4 reads "If there are RATIONAL-valued factors,..." There is a list that most likely isn't too big.

At the end of my last post, I mentioned, "If there are NO Rational Roots, it's a different problem."

This is much more specific than "certain polynomials".

This leads to my original #6.

 

[lookagain]

23851[/ATTACH]
How do I factor it? All I know is how to factor a trinomial by either using PSA 1 or 2, but I'm having a hard time doing this one. Can someone help please? Thank you.

Or, you may see that factor by grouping does not work as is. But, you could ask what
coefficient on \(\displaystyle \ x^2 ,\ \) combined with the last two terms, would give you a trinomial
that you could factor. Here, it would be a 3, so split up the squared term and write it
in a form that you can do some factor by grouping:

\(\displaystyle 2x^3 + 9x^2 + 10x + 3 \ = \)

\(\displaystyle (2x^3 + 6x^2) \ + \ (3x^2 + 10x + 3) \ = \)

\(\displaystyle 2x^2(x + 3) \ + \ (3x + 1)(x + 3) \ = \)

\(\displaystyle (x + 3)(2x^2 + 3x + 1) \ = \)


And continue . . .

 

[Sonal7]

I have this polynomial: View attachment 23851
How do I factor it? All I know is how to factor a trinomial by either using PSA 1 or 2, but I'm having a hard time doing this one. Can someone help please? Thank you.

You know the factor theorem right? Look that up. its useful for these type of Qs. Just checking if you know the fundamental theory of algebra. Gauss proved this which is not something you need to know but this FTA gives you the number of roots that the polynomial has.