Factoring cubics: a^3 - 8b^3, 2x^3 + 9x^2 + 3x - 4

[fred2028]

Hi we're going to learn how to factor cubics this year (grade 11, Ontario), and I just want a head start. I have this worksheet and it has some cubic questions on it. How would you complete these?

a^3 - 8b^3

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

There are other problems but they're related, so I won't post them up. For who ever is going to reply, can you also explain your steps, like don't just do it? Thanks

 

[soroban]

Re: Factoring cubics ...

Hello, Fred!

\(\displaystyle a^3\,-\,8b^3\)

There are two formulas you are expected to know:

\(\displaystyle \begin{array}{cc}\text{Sum of cubes: } \\ \text{Difference of cubes: }\end{array}\;\begin{array}{cc}A^3\,+\,B^3\:=\A\,+\,B)(A^2\,-\,AB\,+\,B^2) \\ A^3\,-\,B^3\:=\A\,-\,B)(A^2\,+\,AB\,+\,B^2)\end{array}\)

Our problem has: \(\displaystyle \,(a)^3\,-\,(2b)^3\)

. . which factors into: \(\displaystyle \,(a\,-\,2b)(a^2\,+\,2ab\,+\,4b^2)\)

\(\displaystyle 2x^3\,+\,9x^2\,+\,3x\,-\,4\)

This one requires the "factor theorem".

If \(\displaystyle x\,=\,a\) makes the polynomial equal 0, then \(\displaystyle (x\,-\,a)\) is a factor.
. . That is, if \(\displaystyle a\) is a "zero" of the polynomial, then \(\displaystyle (x\,-\,a)\) is a factor.
(Note the opposite sign.)

Try some values of \(\displaystyle x.\;\) **

If \(\displaystyle x\,=\,1:\;\;2\cdot1^3\,+\,9\cdot1^2\,+\,3\cdot1\,-\,4\:\neq\:0\) . . . no!

If \(\displaystyle x\,=\,-1:\;\;2(-1)^3\,+\,9(-1)^2\,+\,3(-1)\,-\,4\:=\;0\) . . . yes!

Hence: \(\displaystyle \,(x\,+\,1)\) is a factor.


Using Long Division, we have: \(\displaystyle \,(x\,+\,1)(2x^2\,+\,7x\,-\,4)\)

. . which factors further: \(\displaystyle \,(x\,+\,1)(x\,+\,4)(2x\,-\,1)\;\) . . . There!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

We don't have to try zillions of values for \(\displaystyle x.\)
. . Another theorem narrows down the choices.

Example: \(\displaystyle \,2x^3\,-\,6x\,-\,4x\,+\,3\)

If there are any rational zeros for this polynomial, they must of the form \(\displaystyle \frac{n}{d}\)
. . where \(\displaystyle n\) is a factor of the constant term (3)
. . and \(\displaystyle d\) is a factor of the leading coefficient (2).

The factors of the constant term (3) are: \(\displaystyle \,\pm1,\;\pm3\)

The factor of the leading coefficient are: \(\displaystyle \,\pm1,\;\pm2\)

So the only choices are: \(\displaystyle \,x\:=\:1,\;-1,\;3,\;-3,\;\frac{1}{2},\;-\frac{1}{2},\;\frac{3}{2},\;-\frac{3}{2}\)

And we must test up to eight trial values.
But that's better than the zillions of choices we had a moment ago, right?

 

[Denis]

fred, you can use http://www.google.com (enter "factoring cubics") and
gets lots of helpful sites :idea:

Which Ottawa H.S. do you attend?

 

[fred2028]

fred, you can use http://www.google.com (enter "factoring cubics") and
gets lots of helpful sites :idea:

Which Ottawa H.S. do you attend?

I attend Lisgar Collegiate Institute. You?

 

[Denis]

I attend Lisgar Collegiate Institute. You?

Oh, I'm an old retired guy who loves math...
Went to Glengarry District High School (Alexandria, Ontario);
in the days of "don't spare the rod" :shock:
...when Math teachers treated trigonometric tables the same way the
parish priest treated the bible :wink:

 

[fred2028]

Re: Factoring cubics ...

Hello, Fred!

\(\displaystyle a^3\,-\,8b^3\)

There are two formulas you are expected to know:

\(\displaystyle \begin{array}{cc}\text{Sum of cubes: } \\ \text{Difference of cubes: }\end{array}\;\begin{array}{cc}A^3\,+\,B^3\:=\A\,+\,B)(A^2\,-\,AB\,+\,B^2) \\ A^3\,-\,B^3\:=\A\,-\,B)(A^2\,+\,AB\,+\,B^2)\end{array}\)

Our problem has: \(\displaystyle \,(a)^3\,-\,(2b)^3\)

. . which factors into: \(\displaystyle \,(a\,-\,2b)(a^2\,+\,2ab\,+\,4b^2)\)


[quote:2wo4cji6]\(\displaystyle 2x^3\,+\,9x^2\,+\,3x\,-\,4\)

This one requires the "factor theorem".

If \(\displaystyle x\,=\,a\) makes the polynomial equal 0, then \(\displaystyle (x\,-\,a)\) is a factor.
. . That is, if \(\displaystyle a\) is a "zero" of the polynomial, then \(\displaystyle (x\,-\,a)\) is a factor.
(Note the opposite sign.)

Try some values of \(\displaystyle x.\;\) **

If \(\displaystyle x\,=\,1:\;\;2\cdot1^3\,+\,9\cdot1^2\,+\,3\cdot1\,-\,4\:\neq\:0\) . . . no!

If \(\displaystyle x\,=\,-1:\;\;2(-1)^3\,+\,9(-1)^2\,+\,3(-1)\,-\,4\:=\;0\) . . . yes!

Hence: \(\displaystyle \,(x\,+\,1)\) is a factor.


Using Long Division, we have: \(\displaystyle \,(x\,+\,1)(2x^2\,+\,7x\,-\,4)\)

. . which factors further: \(\displaystyle \,(x\,+\,1)(x\,+\,4)(2x\,-\,1)\;\) . . . There!

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

We don't have to try zillions of values for \(\displaystyle x.\)
. . Another theorem narrows down the choices.

Example: \(\displaystyle \,2x^3\,-\,6x\,-\,4x\,+\,3\)

If there are any rational zeros for this polynomial, they must of the form \(\displaystyle \frac{n}{d}\)
. . where \(\displaystyle n\) is a factor of the constant term (3)
. . and \(\displaystyle d\) is a factor of the leading coefficient (2).

The factors of the constant term (3) are: \(\displaystyle \,\pm1,\;\pm3\)

The factor of the leading coefficient are: \(\displaystyle \,\pm1,\;\pm2\)

So the only choices are: \(\displaystyle \,x\:=\:1,\;-1,\;3,\;-3,\;\frac{1}{2},\;-\frac{1}{2},\;\frac{3}{2},\;-\frac{3}{2}\)

And we must test up to eight trial values.
But that's better than the zillions of choices we had a moment ago, right?

[/quote:2wo4cji6]

Thanks a bunch! I memorized those 2 formulae and now I'm just gonna spend some time on the theorem. Thanks again!