Question

.

Answer 1

\[\text{ Wanting to solve cubics of the form } \\ a^2x^3-b^2x^2-d^2 , a \neq 0 \\a^2x^2(x-\frac{b^2}{a^2})-d^2(x-\frac{b^2}{a^2}-x+\frac{b^2}{a^2}+1) \\ a^2x^2(x-\frac{b^2}{a^2})-d^2(x-\frac{b^2}{a^2})+d^2(x-\frac{b^2}{a^2}-1) \\(x-\frac{b^2}{a^2})(a^2x^2-d^2)+d^2(x-\frac{b^2+a^2}{a^2})

\\
(x-\frac{b^2}{a^2})(ax-d)(ax+d)+\frac{d^2}{a}(ax-\frac{b^2+a^2}{a})\\

\text{ We want } d \text{ to be equal to } \frac{b^2+a^2}{a}\\

\text{ If we have this this then we can write }
\\
(ax-d)[(x-\frac{b^2}{a^2})(ax+d)+\frac{d^2}{a}]\\

(ax-d)(ax^2+dx-\frac{b^2}{a}x-\frac{b^2d}{a^2}+\frac{d^2}{a})\\

(ax-d)(ax^2+(d-\frac{b^2}{a})x-\frac{d}{a}(\frac{b^2}{a}-d))\\
(ax-d)(ax^2+(d-\frac{b^2}{a})x-\frac{d}{a}(\frac{b^2}{a}-d))\\
(ax-d)(ax^2+ax+d) \text{ This should work for cubics satisfying } \\
c=0 \text{ and } d=\frac{b^2+a^2}{a}\\
\\ \text{ Check } \\
a^2x^3+a^2x^2+dax-dax^2-dax-d^2 \\
a^2x^3+x^2(a^2-da)-d^2 \\
a^2x^3+x^2(a^2-(b^2+a^2))-d^2\\
a^2x^3+x^2(-b^2)-d^2\\
a^2x^3-b^2x^2-d^2\\
\text{ Example } \\
\text{ We know } x^3-x^2-4 \text{ is factorable }\\
a^2=1,b^2=1, and d^2=4\\
\text{ Is our condition satisfied ? } \\
4=d^2=(\frac{b^2+a^2}{a})^2=(\frac{1+1}{1})^2=2^2=4 \\
\text{ so the condition does hold }\\
\text{ the factored form of } x^3-x^2+4 \text{ is } \\
(x-2)(x^2+x+2)\\
\text{ Check } \\
x^3+x^2+2x-2x^2-2x-4\\
x^3-x^2-4\\
\text{ Done }\]

Answer 2

@Hero , I provided an example for each one I have a general form for.

Answer 3

I'll have to study this more in depth.

Answer 4

\[a^2x^3-b^2x-d, a \neq 0 \\
a^2x(x^2-\frac{b^2}{a^2})-d \\ a^2x(x-\frac{b}{a})(x+\frac{b}{a})-d(x-\frac{b}{a}+\frac{b}{a}-x+1) \\
a^2x(x-\frac{b}{a})(x+\frac{b}{a} )-d(x-\frac{b}{a})+d(x-\frac{b}{a}-1) \\
a^2x(x-\frac{b}{a})(x+\frac{b}{a})-d(x-\frac{b}{a})+d(x-\frac{b+a}{a}) \\
(x-\frac{b}{a})(a^2x(x+\frac{b}{a})-d)+d(x-\frac{b+a}{a}) \\
(x-\frac{b}{a})(a^2x^2+abx-d)+d(x-\frac{b+a}{a})\\
\text{ Let's factor that one quadratric. } \\
B=(\frac{ab}{2}-z)+(\frac{ab}{2}+z) \text{ and } A \cdot C=-a^2d=(\frac{ab}{2}-z)(\frac{ab}{2}+z)\\
\text{ So }\\
-a^2d=\frac{a^2b^2}{4}-z^2 => z^2-a^2d=\frac{a^2b^2}{4} => z^2=\frac{a^2b^2}{4}+a^2d \\
z=\sqrt{\frac{a^2b^2+4a^2d}{4}}=\frac{a \sqrt{b^2+4d}}{2}\\
\text{ So we have }\\
a^2x^2+abx-d \\ =a^2x^2+(\frac{ab}{2}-\frac{a \sqrt{b^2+4d}}{2})x+(\frac{ab}{2}+\frac{a \sqrt{b^2+4d}}{2})x-d \\ =a^2x(x+\frac{ab-a \sqrt{b^2+4d}}{2a^2})+\frac{ab+a \sqrt{b^2+4d}}{2}(x-\frac{d \cdot 2}{ab+a \sqrt{b^2+4d}})\\
\text{ Then } \\
a^2x^3-b^2x-d \\ = (x-\frac{b}{a})(a^2x^2+abx-d)+d(x-\frac{b+a}{a})\\
=a(x-\frac{b}{a})(x+\frac{b-\sqrt{b^2+4d}}{2a})(ax+\frac{b+\sqrt{b^2+4d}}{2})+d(x-\frac{b+a}{a})\\
\text{ It looks like if } \\
\frac{b-\sqrt{b^2+4d}}{2a}=\frac{-(b+a)}{a}\\
\text{ then we can factor this...}\\
=(x-\frac{b+a}{a})(a(x-\frac{b}{a})(ax+\frac{b+\sqrt{b^2+4d}}{2})+d)\\
=(x-\frac{b+a}{a})((ax-b)(ax+\frac{b+\sqrt{b^2+4d}}{2})+d)\\
=(x-\frac{b+a}{a})(a^2x^2+ax \frac{b+\sqrt{b^2+4d}}{2}-abx-b \frac{b+\sqrt{b^2+4d}}{2}+d)\\
=(x-\frac{b+a}{a})(a^2x^2+(a \frac{b+\sqrt{b^2+4d}}{2}-ab)x+d-b \frac{b+\sqrt{b^2+4d}}{2})\\
\text{ Now lets see if we can simplify that one condition }\\
\frac{b-\sqrt{b^2+4d}}{2}=-(b+a)\\
\frac{-\sqrt{b^2+4d}}{2}=-\frac{2(b+a)+b}{2}\\
\sqrt{b^2+4d}=3b+2a\\
b^2+4d=(3b+2a)^2\\
b^2+4d=9b^2+12ba+4a^2\\
0=8b^2+12ab+4a^2-4d\\
4d=8b^2+8ab+4ab+4a^2\\
4d=8b(b+a)+4a(b+a)\\
d=2b(b+a)+a(b+a)\\
d=(b+a)(2b+a)\\
\text{ So we should be able to factor cubics of the form : } \\
a^2x^3-b^2x-d, a \neq 0 \\
\text{ with the condition }\\
d=(b+a)(2b+a)\\
\text{ The factored form looks like } \\

=(x-\frac{b+a}{a})(a^2x^2+(a \frac{b+\sqrt{b^2+4d}}{2}-ab)x+d-b \frac{b+\sqrt{b^2+4d}}{2})\\
\text{ We know } x^3-x-6 \\
a^2=1,b^2=1,d=6\\
\text{ But does our condition hold }\\
6=d=(1+1)(2 \cdot 1 +1)=2(3)=6\\
\text{ So the answer to that question is yes }\\
\text{ Our factored form for this cubic would be } \\
(x-2)(x^2+( \frac{1+\sqrt{25}}{2}-1)x+6- \frac{1+\sqrt{25}}{2})\\
(x-2)(x^2+2x+3)\\

\text{ Check } \\
x^3+2x^2+3x-2x^2-4x-6\\
x^3-x-6\\
\text{ Done } \\
\]

Answer 5

yep
i just put it on a line

Answer 6

new line*

Answer 7

I recently factored one polynomial and I realize I couldn't have done it without guessing.

x^3 - 14x^2 + 48x - 165

If you can demonstrate that using your general methods, it would be impressive.

Answer 8

Could you factor that just so I can see your steps?

Answer 9

This is the factorization that uses my previous method, but I have since come up with a different method for factoring these:

x^3 - 14x^2 + 48x - 165 = x^2(x - 14) - 3(-16x + 55)
= x^2(x - 14) - 3(x - 14 - 17x + 69)
= x^2(x - 14) - 3(x - 14) + 51x - 207
= x^2(x - 14) - 121(x - 14) +118(x - 14) + 51x - 207
= (x - 14)(x + 11)(x - 11) + 118x - 1652 + 51x - 207
= (x - 14)(x + 11)(x - 11) + 169x - 1859
= (x - 14)(x + 11)(x - 11) + 169(x - 11)
= (x - 11)((x - 14)(x + 11) + 169)
= (x - 11)((x^2 - 3x - 154 + 169)
= (x - 11)(x^2 - 3x + 15)

Answer 10

Since it appears that I have to guess at least one factor x - 11, then the factorization becomes:

x^3 - 14x^2 + 48x - 165 = x^3 - 11x^2 - 3x^2 + 48x - 165
= x^2(x - 11) - 3x^2 + 33x + 15x - 165
= x^2(x - 11) -3x(x - 11) + 15(x - 11)
= (x - 11)(x^2 - 3x + 15)

Answer 11

^That is a method that would work for any factorable polynomial as long as you're able to guess the first factor. Because you can always split the polynomial in accordance with the given factor.

Answer 12

That's the best I've been able to come up with for a consistent method unless you can think of something better.

Answer 13

so we have 48
48=2^4*3
But the only one of these factors that divide 165 is 3

Answer 14

maybe if we have \[ax^3+bx^2+cx+d\]
we have to find a factor of c that divides d
and then factors that out of the cx+d part

Answer 15

Now we have to think about the -3=-121+118
part

Answer 16

you did that to get a perfect square

Answer 17

I figured out the factor because I knew that 11*15 = 165. 11 is prime. For some reason, every factor that has worked so far tends to be prime.

Answer 18

I convinced myself that it doesn't seem to be possible to factor ax^3 + bx^2 + cx + d without guessing at least one factor.