How to refactor a polynomial?

[Fysicus]

Might be a bit basic, but I'm more or less starting from scratch.
We start with the following inequality:

[math]x² + x + 1 > 7[/math]
From what I can remember from my high school days, we solve this by refactoring it, and from what I can find in my old textbooks, the solution can be found like this:

[math]x² - sx + p = (x + x1) (x - x2)[/math]
Where:

[math]x1 + x2 = s[/math][math]x1 . x2 = p[/math]
My question... I can find solutions within my head, but I was wondering... how do we solve those 2 equations?

[math]x1 + x2 = s[/math][math]x1 . x2 = p[/math]

 

[Otis]

\(\displaystyle x^{2} - sx + p = (x + x_1) (x - x_2)\)

\(\displaystyle x_1 + x_2 = s\)
\(\displaystyle x_1 \cdot x_2 = p\)
… how do we solve those 2 equations?

Hi. The generalized setup above is correct for factoring the "easiest" quadratic polynomials, but, since we don't know the signs of s and x₂ (in general), I would change those subtractions to additions.

How we find the numbers x₁ and x₂ depends on the values of p and s. If the polynomial factors nicely (see notes below), then we find x₁ and x₂ in our head or use scratch paper/calculator. Otherwise, we skip factoring and use a different method — to obtain the polynomial's roots — such as numerical approximation methods, quadratic formula, completing the square or graphing.

Before trying to factor, the first step in your exercise is to get zero on the right. Subtract 7 from each side.

x^2 + x – 6 > 0

This polynomial factors nicely. You're looking for two numbers (x₁ and x₂) whose sum is 1 and whose product is -6. Use your knowledge of the multiplication table (and arithmetic with signed numbers) to work it out.

---

Notes:

ax^2 + bx + c = 0

Quadratic polynomials factor nicely when the 'discriminant' is a perfect square.

The discriminant is b^2 – 4ac

For example: x^2 + x – 6
Discriminant: 1^2 – (4)(1)(-6) = 25

For quadratic polynomials that factor nicely: when a=1, we factor using the method above; when a=-1, we multiply each side by -1 first then factor; with other values of a, we use a method called 'factor by grouping'.

 

[blamocur]

Not sure why you need refactoring since there is classical formula for solving quadratic equations.

 

[Dr.Peterson]

My question... I can find solutions within my head, but I was wondering... how do we solve those 2 equations?

[math]x1 + x2 = s[/math][math]x1 . x2 = p[/math]

If you can't find the numbers quickly in your head, you don't bother.

Details are given here:

Factoring Quadratics

Fool-Proof Method for Factoring Simple Quadratics

To factor quadratics in the form x²+bx+c, find factors of c that add up to b. For instance, x²+3x+2 factors as (x+2)(x+1) because 2=2×1 and 3=2+1.

www.purplemath.com

I tend to give myself 10 or 15 seconds to decide whether a quadratic will be easy to factor, and then move to the quadratic formula if I think it will be easier; the risk in the latter is that it isn't self-checking (that is, you can't see at each step if you've made a silly mistake).

 

[blamocur]

the risk in the latter is that it isn't self-checking (that is, you can't see at each step if you've made a silly mistake).

Still, the final results are easy to verify by checking the sum and the product of roots.

 

[Pavlov]

x^2 + x + 1 > 7 subtract '7' from both sides to get
x^2 + x - 6 > 0 factor L side....'in your head' or use Quadratic Formula with a = 1 b = 1 c = -6 to find the roots
(x+3)(x-2) > 0 If BOTH of these factors are negative OR BOTH are positive this is true .... Can you take it from here? Need MORE help?

 

[Otis]

If you can't find the numbers quickly in your head, you don't bother.

This is true most of the time, for me. But sometimes, I will spend a minute or two using paper and/or a calculator.

The OP has not yet responded, but maybe some future readers will appreciate an example.

x^2 + 28x – 533

The discriminant (b^2–4ac) is 2916. Using a calculator, we find that 2916 is a perfect square because [imath]\sqrt{2916} = 54[/imath].

Therefore, this polynomial factors nicely. We need two factors of -533 that sum to 28.

Using basic rules and a calculator, we find the prime factorization of 533 is 13×41. We see that 41–13 is 28, so the two factors of -533 are -13 and 41.

(x – 13)(x + 41) = x^2 + 28x - 533

[imath]\;[/imath]

 

[khansaheb]

.......The discriminant (b^2–4ac) is 2916.

Since you are using the "discriminant" - why not use the quadratic formula and be done quickly.......

 

[Otis]

why not use the quadratic formula and be done quickly

There are a few reasons (eg: showing a different way, reinforcing my mental skills, having fun). I'm not sure what you're thinking in terms of "quickly". Comparing effort for steps taken in post #7 versus steps using the quadratic formula, I don't find much difference.

 

[khansaheb]

Using basic rules and a calculator, we find the prime factorization of 533 is 13×41.

This step can be time consuming.......

 

[Otis]

This step can be time consuming.......

I agree that it might, but not in cases like my example. I only had to check 7 and 13 (in total, the prime factorization required less than 15 seconds).

[imath]\;[/imath]