Understanding Quadratic Equations: Factoring and Finding X-Intercepts

[Caldus]

I just can't factor anything...

I have this equation:

y = -3x^2 - 2x + 5

How do I go about factoring this like:

y = (x + constant)(x + constant)

I tried doing this:

0 = -3x^2 - 2x + 5
(0)-1/3 = (-3x^2 - 2x + 5)-1/3
0 = x^2 + 2/3 - 5/3

I can't factor this:

x^2 + 2/3 - 5/3

(Trying to find the x-intercepts of this equation.)

Thank you.

 

[Hurkyl]

I have this equation:

y = -3x^2 - 2x + 5

How do I go about factoring this like:

y = (x + constant)(x + constant)

You don't!

The work you did is valid, but you encountered the problem of trying to factor with rational coefficients which is difficult... but the final factored form with this approach looks like:

y = (-3)(x + constant)(x + constant)


Anyways, the (generally) easier way is to use the more general factored form:

y = (constant * x + constant)(constant * x + constant)

where all the 'constant's are integers.


And finally, if all else fails, there's always the quadratic formula.

 

[Gale]

since we're on the subject

I started taking an algebra 1 course but 2/3 the way through, i moved to an area that had integrated math. before moving, we were doing factoring of trinomials. I can factor trinomials easily now becaue we spent a lot of time on them. I can't factor any polynomial to higher power than two though. I'm now in calc, and it'd be nice if i could factor things a little better. Is there some general way to know how to factor any polynomial to any degree?

 

[Hurkyl]

One trick is to search for roots.

Theorem:
Suppose f(x) is an integer polynomial (all of its coefficients are integer). If p/q (in lowest terms) is a root of f(x), then p divides the constant term and q divides the leading coefficient.


For example, if p/q is a root of f(x) = 3x³ + 13x² + 19x + 5, then p divides 5 and q divides 3.

The only possibilities for p are 1, 5, -1, -5
The only possibilities for q are 1, 3, -1, -3

So the only possible roots are:

1/1
1/3
-1/1
-1/3
5/1
5/3
-5/1
-5/1

After plugging all of these in, you find that -1/3 works, so (3x + 1) is a factor of f(x).

 

[rdecker]

Originally posted by Hurkyl

So the only possible roots are:

1/1
1/3
-1/1
-1/3
5/1
5/3
-5/1
-5/1

After plugging all of these in, you find that -1/3 works, so (3x + 1) is a factor of f(x). [/B]

Shouldn't that be "the only possible rational roots are"

 

[Hurkyl]

Yes, yes it should.

 

[Asimir]

here's how to solve quadratics that don't have only 1 x^2 (without using the quadratic formula)

find a, b and c using this:
ax^2 + bx + c

so, in your quadratic:
a = -3
b = -2
c = 5

multiply a and c

-3 * 5 = -15

now you need to find two numbers that add to get b and multiply to get a*c
in your case, this is two numbers which add to -2 and multiply to -15
these numbers are 3 and -5

replace the bx part of your equation with these two numbers (as amounts of x)

in your case:

y = -3x^2 + 3x - 5x + 5

now you can factorise it regularly

y = -3x(x - 1) - 5(x - 1)
y = (-3x - 5)(x - 1)

 

[quantumdude]

Caldus,

I'm going to give you a procedure for factoring quadratics that I taught to my College Algebra and Trigonometry class just last week, and I'll use your example as we go along.

Suppose you have a quadratic ax²+bx+c.

Step 0: Check to see if the quadratic is factorable.
To do this, check to see if the discriminant (b²-4ac) is a perfect square. If it is, then you can proceed. If not, then use the quadratic formula.

Let's check your example:

b²-4ac=(-2)²-4(-3)(5)=64

Since 64 is a perfect square, we're in business.

Step 1: Compute the product a×c.

In your problem, a=-3 and c=5, so a×c=-15.

Step 2: Factor a×c such that the factors add up to b.

You have a×c=-15 and b=-2. The factors of -15 that add up to -2 are -5 and 3.

Step 3: Rewrite the quadratic, splitting the middle term according to the factors found in Step 2.

We would rewrite your quadratic as:

-3x²-5x+3x+5

Step 4: Factor by grouping.

In your problem,

(-3x²-5x)+(3x+5)=-x(3x+5)+(3x+5)

Step 5: Bring out the factor common to both terms (there always will be one).

In your problem, the common factor is (3x+5), so:

(3x+5)(-x+1)

is the complete factorization.

edit: fixed bold font bracket

 

[Hurkyl]

Just IMHO: if it takes you more than a few seconds to factor a quadratic equation, you're probably better off using the quadratic formula to find the roots.

 

[rdecker]

quadratic

Originally posted by Hurkyl
Just IMHO: if it takes you more than a few seconds to factor a quadratic equation, you're probably better off using the quadratic formula to find the roots.

You're quite right. If your goal is to find the roots then it is probably easier to program the quadratic formula into your calculator and enter a,b and c. I teach math in a high school and I get my students to learn how to write the program early in the course. However, I appreciate the process of factoring as mental gymnastics. I also use factoring to arrive at a factored equation which I can use to teach the properties of zero products.

 

[Big Jeff]

When I need to solve a quadratic, I simply solve it by completing the square (unless, of course, it's already a perfect square trinomial). No formula to memorize, no factoring, just the same procedure every time.

 

[rdecker]

yes, but

Completing the square gives you the vertex form but the roots are not immediately solved.

 

[Big Jeff]

Agreed.

 

[quantumdude]

Originally posted by Hurkyl
Just IMHO: if it takes you more than a few seconds to factor a quadratic equation, you're probably better off using the quadratic formula to find the roots.

I'm assuming that Caldus is at the same point in his course that my students are in mine. If so, then he has not seen the quadratic formula yet.

 

[IceZero]

All I can say to you, is to learn synthetic division my friend, and know the formula for addition and subtraction of perfect cubes. Synthetic division makes your life heaven when it comes to polynomials of the 3rd,4th,5th... degree.

 

[Pauly Man]

Originally posted by Tom
I'm assuming that Caldus is at the same point in his course that my students are in mine. If so, then he has not seen the quadratic formula yet.

Are college students equivalent to high school students or uni students or in between? (An unfamiliar concept to me, being from Aus).

Anyway we learned the quadratic formula back at the beginning of high school, when are US students introduced to it?

 

[quantumdude]

Originally posted by Pauly Man
Are college students equivalent to high school students or uni students or in between? (An unfamiliar concept to me, being from Aus).

A college student is an undergraduate at a university.

Anyway we learned the quadratic formula back at the beginning of high school, when are US students introduced to it?

There's a wide variety of US high school students. I learned the quadratic formula in 8th grade, whereas some do not learn it at all.

 

[Gale]

in public school in the US the quadratic is learned in Math2 if its intgrated math, and algebra 1 if its the old way. That's usually freshman or sophmore year of high school. that's the average, advance student learn sooner, slower learn later obviously.