Factoring Polynomials: A Faster Way!

[paulmdrdo1]

Hello!

I just want to know the faster way to factor the expression. I already factored it out using trial and I am hoping you could give me some tricks to go about it faster than the usual method. Thanks

$192x^3-164x^2-270x$

 

[MarkFL]

We see that $2x$ is a factor of all 3 terms, so we may write:

\(\displaystyle 2x(96x^2-82x-135)\)

Now, on the quadratic factor, I would use the quadratic formula to determine the roots:

\(\displaystyle x=-\frac{5}{6},\,\frac{27}{16}\)

And since \(\displaystyle 96=6\cdot16\), we can write:

\(\displaystyle 2x(6x+5)(16x-27)\)

 

[Deveno]

Factoring polynomials is a process fraught with peril, it might be that all our attempts end in failure.

If one assumes (because maybe your instructor has told you so) that a polynomial *does* indeed have a factorization, we can use a tool called "the rational root test", which tells us that any (non-zero) root (and there might not be any) in the rationals is of the form:

$\pm\dfrac{p}{q}$

where $p$ is an integer that divides 135, and $q$ is an integer that divides 96 (because of MarkFL's factoring out the $2x$).

This still leaves several possibilities to be checked out with "trial and error", but it at least cuts down on the sheer number of tries.

The long and short of this is:

with a polynomial of degree 3, "guess and check" for the rational roots is probably the fastest (the "formula" for cubics is pretty nasty),

with a polynomial of degree 2, the quadratic formula is usually your friend (unless you can "see" the factorization at once).

Factoring polynomials of degree 4 is often *very* difficult (unless there's no $x^3$ and $x$ terms), and factoring polynomials of degree 5 is often a lost cause.

 

[paulmdrdo1]

Hello!

From what MarkFl provided, he used the quadratic formula to find the roots of $96x^2-82x-135$
Since the roots are x = -5/6 and x = 27/16 I am thinking that the factored form of $96x^2-82x-135$
Would be $(x+\frac{5}{6})(x-\frac{27}{16}) = (\frac{(6x+5)(16x-27)}{96})$
So the completely factored form of the entire expression could be

$(\frac{(6x+5)(16x-27)x}{48})$

I know it is wrong but can you tell me why?

 

[MarkFL]

You are correct when you state:

\(\displaystyle \left(x+\frac{5}{6}\right)\left(x-\frac{27}{16}\right)=\frac{(6x+5)(16x-27)}{96}\)

However, the LHS is equal to:

\(\displaystyle \frac{96x^2-82x-135}{96}\)

Therefore:

\(\displaystyle 96x^2-82x-135=(6x+5)(16x-27)\)

You see for any roots of a quadratic (or any polynomial), there are an infinite number of polynomials having those roots. In the case of the quadratic, let's say it has the roots $r_1,\,r_2$. Then the family of quadratics having those two roots is given by:

\(\displaystyle f(x)=k(x-r_1)(x-r_2)\) where \(\displaystyle 0\ne k\)

Does that make sense?

 

[paulmdrdo1]

You are correct when you state:

\(\displaystyle \left(x+\frac{5}{6}\right)\left(x-\frac{27}{16}\right)=\frac{(6x+5)(16x-27)}{96}\)

However, the LHS is equal to:

\(\displaystyle \frac{96x^2-82x-135}{96}\)

Therefore:

\(\displaystyle 96x^2-82x-135=(6x+5)(16x-27)\)

You see for any roots of a quadratic (or any polynomial), there are an infinite number of polynomials having those roots. In the case of the quadratic, let's say it has the roots $r_1,\,r_2$. Then the family of quadratics having those two roots is given by:

\(\displaystyle f(x)=k(x-r_1)(x-r_2)\) where \(\displaystyle 0\ne k\)

Does that make sense?

Hello!
From what I understand when you know the roots of a certain polynomial we can use those roots to get the factored form a that polynomial.
Is the expression \(\displaystyle \left(x+\frac{5}{6}\right)\left(x-\frac{27}{16}\right)\) equivalent to $96x^2-82x-135$? And if it is how do we transform it to $96x^2-82x-135$

Please bear with me.