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Step 8Factoring polynomials by synthetic division
Sometimes you'll get beastly polynomials that look like they have no hope.
3x^3 + 8x^2 - 9x + 2 is an example. You can't use grouping to factor out a GCF in a way that would produce a common factor.
In order to explain how this works, you need to know that when solving an equation by factoring, you need to set the factored out thing equal to 0 and find out what X equals so that it equals zero. For example, 0 = (x - 2) (x + 1). The solutions are 2 and -1.
If a polynomial has integer coefficients, every zero, or solution, has the form P/Q, where P = a factor of the constant term, and Q = a factor of the leading coefficient.
Basically, if you list all the factors of the constant, and divide them by the factors of the leading coefficient (the coefficient next to the variable with the highest power) in every combination, you will get a list of possible rational solutions. How does this help you factor? If you get 2 as a solution, you can work backwards and say that one of the factors of the equation was (x - 2).
So, back to the example:
Factors of 2: +/- 1, +/- 2 (you need to include negatives)
Factors of 3: +/- 1, +/- 3
P/Q: +/- 1, +/- 1/3, +/- 2, +/- 2/3
Once you have your list, you'll use something called synthetic division to see which of those P/Q's are actually solutions.
Synthetic division is a way of dividing polynomials by a binomial of the form x-k. I'm not going to explain how it works, but just show how to use it for factoring.
First, put one of your P/Q's in a little box or set of parentheses, then list the coefficients and constant in a row next to it. If the polynomial skips a power (x^2 + 2) then you need to add a 0 for where x1 should have been.
(Equation: 3x^3 + 8x^2 - 9x + 2)
(Ignore the asterisks, they're used as placeholders. Better yet, see the first picture.)
(1) 3 8 -9 2
Leave a blank space, draw a line, then drop the first term, 3, down.
(1) 3 8 -9 2
***3
Then multiply it by the number in the box and put it beneath the next term.
(1) 3 8 -9 2
******3
***3
Add 8 + 3
(1) 3 8 -9 2
******3
***3 11
Multiply.
(1) 3 8 -9 2
******3 11
***3 11
Add.
(1) 3 8 -9 2
******3 11
***3 11 2
Multiply.
(1) 3 8 -9 2
******3 11 2
***3 11 2
Add.
(1) 3 8 -9 2
******3 11 2
***3 11 2 4
That string of numbers, 3, 11, 2, 4, gives you an equation with one degree less (if the highest exponent in the original equation is 3, the highest exponent in the quotient will be a 2) as well as a remainder.
(Original Equation: 3x^3 + 8x^2 - 9x + 2)
Quotient: 3x^2 + 11x + 2 Remainder 4
If you get a remainder, then the number in the box you tried is not a solution for the equation. Cross that number off your list, and try again with another number. It's pretty much guess and check.
Eventually you'll try 1/3 and you'll find that it divides through cleanly. You'll end up with:
(x - 1/3) (3x^2 + 9x - 6).
Now that you have a trinomial of power two, you can go back and factor it. Don't forget to take out the GCF first! You're left with (x - 1/3) (3) (1x^2 + 3x + 2). Factor out the trinomial via the quadratic equation (this expression was used as an example in step [6], so refer back if you need to). You'll end up with (3) (x - 1/3) (x - ((-3 + sqrt 17)/2)) (x - ((-3 - sqrt 17)/2)). Very ugly, but that's how you do it.
3x^3 + 8x^2 - 9x + 2 is an example. You can't use grouping to factor out a GCF in a way that would produce a common factor.
In order to explain how this works, you need to know that when solving an equation by factoring, you need to set the factored out thing equal to 0 and find out what X equals so that it equals zero. For example, 0 = (x - 2) (x + 1). The solutions are 2 and -1.
If a polynomial has integer coefficients, every zero, or solution, has the form P/Q, where P = a factor of the constant term, and Q = a factor of the leading coefficient.
Basically, if you list all the factors of the constant, and divide them by the factors of the leading coefficient (the coefficient next to the variable with the highest power) in every combination, you will get a list of possible rational solutions. How does this help you factor? If you get 2 as a solution, you can work backwards and say that one of the factors of the equation was (x - 2).
So, back to the example:
Factors of 2: +/- 1, +/- 2 (you need to include negatives)
Factors of 3: +/- 1, +/- 3
P/Q: +/- 1, +/- 1/3, +/- 2, +/- 2/3
Once you have your list, you'll use something called synthetic division to see which of those P/Q's are actually solutions.
Synthetic division is a way of dividing polynomials by a binomial of the form x-k. I'm not going to explain how it works, but just show how to use it for factoring.
First, put one of your P/Q's in a little box or set of parentheses, then list the coefficients and constant in a row next to it. If the polynomial skips a power (x^2 + 2) then you need to add a 0 for where x1 should have been.
(Equation: 3x^3 + 8x^2 - 9x + 2)
(Ignore the asterisks, they're used as placeholders. Better yet, see the first picture.)
(1) 3 8 -9 2
Leave a blank space, draw a line, then drop the first term, 3, down.
(1) 3 8 -9 2
***3
Then multiply it by the number in the box and put it beneath the next term.
(1) 3 8 -9 2
******3
***3
Add 8 + 3
(1) 3 8 -9 2
******3
***3 11
Multiply.
(1) 3 8 -9 2
******3 11
***3 11
Add.
(1) 3 8 -9 2
******3 11
***3 11 2
Multiply.
(1) 3 8 -9 2
******3 11 2
***3 11 2
Add.
(1) 3 8 -9 2
******3 11 2
***3 11 2 4
That string of numbers, 3, 11, 2, 4, gives you an equation with one degree less (if the highest exponent in the original equation is 3, the highest exponent in the quotient will be a 2) as well as a remainder.
(Original Equation: 3x^3 + 8x^2 - 9x + 2)
Quotient: 3x^2 + 11x + 2 Remainder 4
If you get a remainder, then the number in the box you tried is not a solution for the equation. Cross that number off your list, and try again with another number. It's pretty much guess and check.
Eventually you'll try 1/3 and you'll find that it divides through cleanly. You'll end up with:
(x - 1/3) (3x^2 + 9x - 6).
Now that you have a trinomial of power two, you can go back and factor it. Don't forget to take out the GCF first! You're left with (x - 1/3) (3) (1x^2 + 3x + 2). Factor out the trinomial via the quadratic equation (this expression was used as an example in step [6], so refer back if you need to). You'll end up with (3) (x - 1/3) (x - ((-3 + sqrt 17)/2)) (x - ((-3 - sqrt 17)/2)). Very ugly, but that's how you do it.
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