I'm having difficulty factoring trinomials, quadrinomials, a

[warwick]

nd higher degree polynomials. My last Pre-Calculus course was in 2002 in high school. Needless to say, it's been awhile, and I'm rusty. Anyway, I'm looking at my Pre-Calculus book:

Solve (x^4) + (x^3) - 15(x^2) - 3x + 36 < 0

The listed factors in the book are (x - 3) (x + 4) ((x^2) - 3), and the other roots are plus/minus the square root of (3). When I factor by grouping, I don't get the same results.

Here are a few more:

(x^3) - 4(x^2) - 20x + 48 >/equal to 0

f(x) = - 2(x^3) - 8(x^2) + 6x + 36

Of course, it's easy to find the zeroes if the polynomial is in quadratic form. Thanks for any help.

 

[stapel]

When I factor by grouping, I don't get the same results.

Please show the steps you did and the results you got.

Thank you.

Eliz.

 

[warwick]

When I factor by grouping, I don't get the same results.

Please show the steps you did and the results you got.

Thank you.

Eliz.

(x^4) + (x^3) - 15(x^2) - 3x + 36

((x^4) + (x^3) - 15(x^2)) (- 3x + 36)

x^2 (x^2 + x - 15) - 3(x - 12)

(x^2 - 3) (x^2 + x - 15) (x - 12)

From that, I can see from where the plus/minus the square root of 3 comes.

Could you do the other the two and show the steps?

 

[stapel]

Addition and multiplication are not interchangeable. You can't go from your first line to your second line. (An easy check: If you multiplied x⁴ + x³ - 15x² by -3x + 36, you would get -3x⁵ as the leading term, which is not what you'd started with.)

This polynomial does not factor "by grouping", at least not in any obvious way that I can see. I would suggest starting with the Rational Roots Test, to get a list of potential zeroes. Then use synthetic division to find which, if any, work. You should be able to reduce the polynomial to linear factors and, perhaps, a quadratic, to which you can apply the Quadratic Formula if necessary.

Eliz.

 

[warwick]

Addition and multiplication are not interchangeable. You can't go from your first line to your second line. (An easy check: If you multiplied x⁴ + x³ - 15x² by -3x + 36, you would get -3x⁵ as the leading term, which is not what you'd started with.)

This polynomial does not factor "by grouping", at least not in any obvious way that I can see. I would suggest starting with the Rational Roots Test, to get a list of potential zeroes. Then use synthetic division to find which, if any, work. You should be able to reduce the polynomial to linear factors and, perhaps, a quadratic, to which you can apply the Quadratic Formula if necessary.

Eliz.

Then how when can I factor by grouping? I thought I was applying it correctly. Obviously, I did not check myself, though.

Using synthetic division, can I successively divide a polynomial by a zero from nth degree to quadratic form? In other words, for this polynomial, could I synthetically divide it by two different zeros to get it down to quadratic form?

Can you find the solutions to the other two problems in the initial post? Thanks.

 

[galactus]

As Stapel mentioned, are you familiar with the rational root test?.

When a polynomial is not easily grouped this is a method you can use to factor.

It can be somewhat laborious if you have to test the roots by hand.

Personally, I'd use a calculator to test them. Unless of course, you want to practice

your synthetic division.


Yes, after you find a root(s) you can use them to reduce the polynomial to a

quadratic.

For the first one, \(\displaystyle x^{4}+x^{3}-15x^{2}-3x+36\)

Using the Rational Root test, we find that 3 is a root.

Now, divide the expression by (x-3). See what you get.....betcha it reduces.

Find another root and it'll reduce further.....down to your quadratic.

That will be four roots, as the First Fundamental Theorem of Algebra so

says.

 

[warwick]

As Stapel mentioned, are you familiar with the rational root test?.

When a polynomial is not easily grouped this is a method you can use to factor.

It can be somewhat laborious if you have to test the roots by hand.

Personally, I'd use a calculator to test them. Unless of course, you want to practice

your synthetic division.


Yes, after you find a root(s) you can use them to reduce the polynomial to a

quadratic.

For the first one, \(\displaystyle x^{4}+x^{3}-15x^{2}-3x+36\)

Using the Rational Root test, we find that 3 is a root.

Now, divide the expression by (x-3). See what you get.....betcha it reduces.

Find another root and it'll reduce further.....down to your quadratic.

That will be four roots, as the First Fundamental Theorem of Algebra so

says.

I am familiar with the rational root test, but sometimes there quite a few roots, with some being in unpretty rational forms, and two imaginary roots. Of course, Descartes change of sign rule will give me the information necessary to determine the nature of the roots.

 

[stapel]

You might want to review the topic of "factoring by grouping", as well. The technique depends upon the ability to extract common factors, not just any old factors you please. For instance:

. . . . .x³ - 5x² + 3x - 15

. . . . .(x³ - 5x²) + (3x - 15)

. . . . .x²(x - 5) + 3(x - 5)

. . . . .(x - 5)(x² + 3)

Take particular note of the second step: I regrouped, but I didn't make the addition magically disappear.

In the above example, if there had not been the common factor of "x - 5", the "by grouping" factorization would have failed.

This is what you did:

. . . . .x⁴ + x³ - 15x² - 3x + 36

. . . . .(x⁴ + x³ - 15x²)(-3x + 36)

What happened to the sign in between the two parts? This addition has magically become multiplication. The attempt at grouping should have been written as:

. . . . .(x⁴ + x³ - 15x²) + (-3x + 36)

. . . . .x²(x² + x - 15) + (-3)(x - 12)

But there are no common factors. The x² is multiplied by x² + x - 15, and the -3 is multiplied by x - 12, so nothing factors out.

Unfortunately, many instructors use "by factoring" in the special case of quadratics, and tolerate very casual notation, which, as in the above thread, leads later to great confusion.

Instead, try using the Rational Roots Test, and work from there.

Eliz.

 

[galactus]

All the roots are real for this particular problem.

 

[warwick]

You might want to review the topic of "factoring by grouping", as well. The technique depends upon the ability to extract common factors, not just any old factors you please. For instance:

. . . . .x³ - 5x² + 3x - 15

. . . . .(x³ - 5x²) + (3x - 15)

. . . . .x²(x - 5) + 3(x - 5)

. . . . .(x - 5)(x² + 3)

Take particular note of the second step: I regrouped, but I didn't make the addition magically disappear.

In the above example, if there had not been the common factor of "x - 5", the "by grouping" factorization would have failed.

This is what you did:

. . . . .x⁴ + x³ - 15x² - 3x + 36

. . . . .(x⁴ + x³ - 15x²)(-3x + 36)

What happened to the sign in between the two parts? This addition has magically become multiplication. The attempt at grouping should have been written as:

. . . . .(x⁴ + x³ - 15x²) + (-3x + 36)

. . . . .x²(x² + x - 15) + (-3)(x - 12)

But there are no common factors. The x² is multiplied by x² + x - 15, and the -3 is multiplied by x - 12, so nothing factors out.

Unfortunately, many instructors use "by factoring" in the special case of quadratics, and tolerate very casual notation, which, as in the above thread, leads later to great confusion.

Instead, try using the Rational Roots Test, and work from there.

Eliz.

So, use the rational roots test and trial and error? Sometimes, as I said, the RRT yields some nasty quotients.

 

[stapel]

I would suggest starting with the Rational Roots Test....

...are you familiar with the rational root test? When a polynomial is not easily grouped this is a method you can use to factor.

...try using the Rational Roots Test, and work from there.

So, use the rational roots test and trial and error?

Yes; use the Rational Roots Test.

Using the Rational Root test, we find that 3 is a root....All the roots are real for this particular problem.

Sometimes...the RRT yields some nasty quotients.

Yes, but the roots are all real this time, and, in any case, "grouping" won't work. So use the Rational Roots Test.

Eliz.