# Solving Polynomials: Factoring

1. Jul 27, 2015

### Mary1910

1. The problem statement, all variables and given/known data

Solve for x.

a) 2x^3-3x^2-5x+6=0

2. Relevant equations

-Find possible values of x
-Divide that factor by 2x^3-3x^2-5+6 using long division

3. The attempt at a solution

ƒ(2)=2(2)^3-3(2)^2-5(2)+6
ƒ(2)=16-12-10+6
ƒ(2)=0

Now using long division divide 2x^3-3x^2-5x+6 by (x-2)

This gave me 2x^2+x-3

∴ƒ(x)=(x-2)(2x^2+x-3)

x=2

Im not sure about these types of questions, I know this is probably a simple question to most of you but how do I determine if there is more than one value of x? I used x=-1±√(1)^24(2)(-3)/(2(2)) which resulted in -1±√-19/(4). does this mean that x=2 and the other roots are not real numbers since (√-19).

Thank you, any help would be appreciated :)

2. Jul 27, 2015

### aikismos

Well,

If the QF is $x = {{-b ± \sqrt{b^2 -4ac}}\over{2a}}$, then you shouldn't have a 24 in your radical as part of product. When I do the calculations, I get roots at $1$ and $-{3}\over{2}$, so yes, ∃ 3 distinct real roots. Had it been yours, then the two of the roots would have been imaginary reducing to expression like $x = {{-1 ± i\sqrt{19}}\over{4}}$. Related to that is Descartes' Rule of Signs https://en.wikipedia.org/wiki/Descartes'_rule_of_signs

3. Jul 27, 2015

### tommyxu3

When solving $2x^2+x-3=0,$ you should get $x=\frac{-1\pm\sqrt{1^2-4\cdot 2\cdot (-3)}}{2\cdot 2}=1,\frac{-3}{2},$ or using the factorizing $2x^2+x-3=(2x+3)(x-1)=0\Rightarrow x=1,\frac{-3}{2}.$
I think you may have some mistakes when applying formula to calculate the roots.

4. Jul 27, 2015

### Mary1910

Okay so there are 3 roots and they are x=2, x=1, and x=-3/2. I see where I had made errors before in my calculations. The only part I still don't quite understand is how found x=-3/2 without factoring? Did you just use the value of (c) in x=-b±√b^2-4(a)(c) / 2(a) ??

5. Jul 27, 2015

### tommyxu3

Yes! Mind the sign $\pm$ in it, which would create two values.

6. Jul 27, 2015

### aikismos

But you did factor. The essence of the quadratic formula is that is a pre-factored form of any quadratic equation where the range (or codomain value if you prefer) is set to 0, required by the definition of a root. Remember that the QF is the third and last in a series of factoring techniques; it's the most complicated, but it works without exception as proven by the derivation being algebraically consistent with the axioms of mathematics. For fun:

1. $ax^2 + bx + c = f(x)$
2. $ax^2 + bx + c = 0$
3. $ax^2+bx=-c$
4. $x^2 + {bx \over a } + {b^2 \over 4a^2 }= {b^2 \over 4a^2 } +{-c \over a }$
5. $(x + {b \over 2a } )^2 = {b^2 - 4ac \over 4a^2}$
6. $x + {b \over 2a } = \pm \sqrt{b^2 - 4ac \over 4a^2}$
7. $x = - {b \over 2a } \pm {\sqrt{b^2 - 4ac} \over 2a}$
8. $x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}$
What this really means graphically, is that anytime your parabola crosses the y-axis above the vertex, you always have two distinct real roots that are equidistant from the axis of symmetry that runs through the vertex of the parabola (found at $-b \over 2a$.) The left root will be $( - {b \over 2a } - {\sqrt{b^2 - 4ac} \over 2a}, 0)$ and the right one will be $(- {b \over 2a } + {{\sqrt{b^2 - 4ac} \over 2a}}, 0)$.

It can be written as factors like this:
$(x - (- {b \over 2a } + {{\sqrt{b^2 - 4ac} \over 2a}}))(x - (- {b \over 2a } - {{\sqrt{b^2 - 4ac} \over 2a}})) = 0$
or
$(x + ( {b \over 2a } - {{\sqrt{b^2 - 4ac} \over 2a}}))(x + ( {b \over 2a } + {{\sqrt{b^2 - 4ac} \over 2a}})) = 0$.

7. Jul 27, 2015

### tommyxu3

Yes, that's what I did. I just answer his/her question.

8. Jul 27, 2015

### Mary1910

Hey I have another question about a similar problem if I could get some help that would be great,

1. The problem statement, all variables and given/known data

Solve for x.

8x^3+4x^2-18x-9

2. Relevant equations

find possible values for x

So oddly this is the part that has got me. First I know that there is a max of three zeros. Then I know I need to find the list of pairings equal to -9. Such as 1, -9 and -1, 9 and 3, -3. However after trying to substitute each of these values for x, I still haven't been able to determine a which is a factor. I don't know whether its a "trick question" that can't be answered, there's a type-o, or I whether I have just completely missed something.

Any help would be appreciated, Thanks.
BTW, I don't know if we can post more than one question on the same thread, but given the question is so similar to the original, I just assumed that it would be fine.

9. Jul 27, 2015

### SteamKing

Staff Emeritus
You seem to be trying to solve polynomials in a hit-or-miss fashion.

This particular polynomial is a little different because the coefficient of the leading term ("8") is not equal to 1.

You should examine the Rational Root Theorem to give you some better choices of roots for this polynomial:

https://en.wikipedia.org/wiki/Rational_root_theorem

You might also read up on Descartes Rule of Signs while you're at it:

https://en.wikipedia.org/wiki/Descartes'_rule_of_signs

These two techniques help you examine a given polynomial and determine how many real roots there are and which choices of real roots may be worth examining.

10. Jul 28, 2015

### aikismos

Heya Mary,

Here's another fun polynomial for finding roots (when I say fun, I mean you can use algebra as opposed to numerical techniques :D).

Before we begin though, by using the two pound symbols (#) in a row to open and close your math, you can make it easier for us to read. E.g.:

(#)(#) 8x^3+4x^2-18x-9 (#)(#) -> $8x^3+4x^2-18x-9$ (you can right click on this too, to see it in TeX form)

Anyway, this polynomial can be attacked with a technique called factor by grouping:
1. $8x^3+4x^2-18x-9$ (NB that the pairs of coefficients are multiples 8,4 and 18, 9
2. $4x^2(2x+1)-9(2x+1)$ (NB that we now have a binomial $AC-BC$, where $C$ is itself the binomial
3. $(2x+1)(4x^2-9)$ (NB we've merely changed $AC+BC \rightarrow C(A-B)$)
4. $(2x+1)(2x + 3)(2x-3)$ (NB that the last binomial quadratic is a difference of squares $A^2-B^2 \rightarrow (A+B)(A-B)$
If we were trying to find roots, we'd set this equal to zero and then use the Zero Product Property to solve for each $x$.
1. $(2x+1)(2x + 3)(2x-3)=0$ (NB $ABC=0 \rightarrow A=0 ∧B=0 ∧C=0$)
2. $x = \{-{1 \over 2}, -{3 \over 2}, {3 \over 2}\}$
I'll follow this post up with using this polynomial with the rational root theorem.

Last edited: Jul 28, 2015
11. Jul 28, 2015

### aikismos

Let's take the polynomial $8x^3+4x^2-18x-9$ and do a substitution letting $p=8$ (the leading coefficient) and $q=9$ (the trailing coefficient). We get $px^3+4x^2-18x-q$. The rational root theorem says that the roots must take a particular form: $x=\pm{q_n \over p_m }$ where $p_m, q_n$ are factors of $p,q$ respectively.

To wit, the factors of our respective coefficients are:

1. $q_n = \pm \{1, 3, 9\}$
2. $p_m = \pm \{1, 2, 4, 8\}$
This means that the roots can be among the following: $\{ \pm {1 \over 1}, \pm {3 \over 1}, \pm {8 \over 1}, \pm {1 \over 2}, \pm {3 \over 2}, \pm {8 \over 2}, \pm {1 \over 4} , \pm {3 \over 4}, \pm {8 \over 4} , \pm {1 \over 8}, \pm {3 \over 8}, \pm {8 \over 8}\}$, and this set is equivalent after tidying up our candidate fractions to $\{ \pm {1 \over 8}, \pm {1 \over 4}, \pm {3 \over 8}, \pm {1 \over 2}, \pm {3 \over 4}, \pm 1, \pm {3 \over 2}, \pm 2, \pm 3 \pm 4, \pm 8\}$. Now, you can either solve this by testing each of these and determining which inputs give us an output of zero, or a faster way would be to use software like Geogebra which is free and has a computer-algebra system (CAS) built in it, to graph the polynomial and determine the roots.
Then, where the fun starts is playing around with the graph and seeing how the zeros change as you tinker with the polynomial. What happens if you negate all the coefficients? What happens if you negate just the leading? Usw.

Last edited: Jul 28, 2015
12. Jul 28, 2015

### aikismos

Polynomial graph looks like this:

Close up of the roots looks like this:

It looks like $x=\{-{3 \over 2}, -{1 \over 2}, {3 \over 2}\}$ which is convenient, since above, we found through 'factor by grouping'
$x = \{-{1 \over 2}, -{3 \over 2}, {3 \over 2}\}$ is indeed the answer. :D

So, in summary the rational root theorem, factoring, and the use of the zero product property constitute a generally effective approach for systematically finding roots for textbook polynomials!

Another explanation can be found here http://www.purplemath.com/modules/rtnlroot.htm

13. Jul 28, 2015

### aikismos

Erratum (sorry, responded at lunch):

"This means that the roots can be among the following..."

"This means that the roots can be among the following: $\pm {1 \over 1}, \pm {3 \over 1}, \pm {9 \over 1}, \pm {1 \over 2}, \pm {3 \over 2}, \pm {9 \over 2}, \pm {1 \over 4} , \pm {3 \over 4}, \pm {9 \over 4} , \pm {1 \over 8}, \pm {3 \over 8}, \pm {9 \over 8}\}$ , and this set is equivalent after tidying up our candidate fractions to $\pm {1 \over 8}, \pm {1 \over 4} , \pm {3 \over 8},\pm {1 \over 2}, \pm {3 \over 4}, \pm 1,\pm {9 \over 8}, \pm {3 \over 2}, \pm {9 \over 4} , \pm 3, \pm {9 \over 2}, \pm 9 \}$.