x^3 - 2x^2 - x + 2
What steps are used to factor this?
x^3 - 2x^2 - x + 2
What steps are used to factor this?
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Verified answer
Hi,
x³ - 2x² - x + 2 =
x²(x - 2) - 1(x - 2) =
(x² - 1)(x - 2) =
(x - 1)(x + 1)(x - 2) <==ANSWER
Here are general rules and examples for factoring quadrinomials. Let's look at these 4 term problems. As always, look for a GCF first.
Then try to factor a GCF out of just the first 2 terms and then try to factor a GCF out of the last 2 terms. If you can do this, the expression left behind both times must be the exact same expression. The 2 GCFs go together to make one factor and the repeated expression is the second factor. If either expression has an exponent in it, check to see if it can be factored again.
Given the problem: 5x³ - 10x² +3x - 6
There is no GCF for the entire problem.
The first 2 terms has a GCF of 5x². Factor it out.
5x³ - 10x² + 3x - 6
5x²(x - 2) + 3x - 6
Now factor out the GCF of the last 2 terms.
5x²(x - 2) + 3x - 6
5x²(x - 2) + 3(x - 2)
Now put 5x² and + 3 together as one factor. (x - 2) is the other factor.
(5x² + 3)(x - 2) are the factors. Nothing factors again.
Another problem is 4a³ + 28a² -9a - 63
There is no GCF for the entire problem.
The first 2 terms has a GCF of 4a². Factor it out.
4a²(a + 7) -9a - 63
Now factor out the GCF of the last 2 terms.
4a²(a + 7) -9(a + 7) Note the GCF had to be a -9 so that the parentheses would be the same.
Now put 4a² and - 9 together as one factor. (a + 7) is the other factor.
(4a² - 9)(a + 7) are the factors so far, but 4a² - 9 is the difference of perfect squares. It factors again.
(2a - 3)(2a + 3)(a + 7) These are the final factors.
A second type of 4 term problem is something like this:
9x² + 12xy + 4y² - 16z²
In this problem there is no GCF for all 4 terms. The first 2 terms have a GCF and the last 2 terms have a GCF, but the expressions are not the same, so the 2 term - 2 term method doesn't work.
9x² + 12xy + 4y² - 16z²
3x(3x + 4y) + 4(y² - 4z²) Note parentheses are different, so this doesn't work.
Instead of grouping 2 and 2, try grouping 3 and 1.
9x² + 12xy + 4y²....... - 16z²
Try factoring the first 3 terms as a trinomial, hoping to get a binomial squared.
9x² + 12xy + 4y²....... - 16z²
(3x + 2y)² - 16z²
Now factor this as the difference of perfect squares.
Square (3x + 2y) to get (3x + 2y)² and square 4z to get 16z²
That means the factors both start with 3x + 2y and both end with 4z, with a "+" or "-" in front of 4z.
(3x + 2y)² - 16z²
(3x + 2y - 4z)(3x + 2y + 4z) These are the factors!!
I hope that helps!! :-)
Quadrinomial Factoring
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To factor the bottom, we would need to get it into the form (x + a)(x + b)(x + c). If we multiply that out to see how it corresponds to the bottom in its current state, we get: (x + a)(x + b)(x + c) = (x^2 + (a + b)x + ab)(x + c) = x^3 + (a + b)x^2 + abx + cx^2 + (ac + bc)x + abc = x^3 + (a + b + c)x^2 + (ab + ac + bc)x + abc Comparing that to the bottom gives: a + b + c = -5 abc = 9 So we want three numbers that multiply together to give 9 and add up to -5. A bit of trial and error and considering the factors of 9 shows that the values of a, b and c we want are -3, -3 and 1. So we can now simplify the original expression as: (x^2 + 15)/(x + 1)(x - 3)^2
Sum of the coefficients of all powers = 0
So x-1 is a factor.
Using synthetic division, the other factor is found as x^2 -x -2.
This can be further factorised as (x-2) (x+1)
Hence the three factors of x^3 -2x^2 -x +2 = (x-1)(x-2) (x+1)
x³ - 2x² - x + 2
=> x²( x - 2 ) - x + 2
=> x²( x - 2 ) - (x - 2)
=> ( x² - 1 )( x - 2 )
Note the diff of two squares. Factor that and you'll have it.
:)
Split the quadratic term.
x³ - x² - x² - x + 2
(x³ - x²) + (-x² - x + 2)
x²(x - 1) - (x² + x - 2)
x²(x-1) - (x + 2)(x-1)
(x-1)(x² - (x+2))
(x-1)(x² - x - 2)
(x - 1)(x - 2)(x + 1)
(x - 1)(x + 1)(x - 2)
That one was easy; simply factor by grouping. Others well, viete's formulas will work for just about anything.