To factor 8x^3 + 4x^2 - 2x - 1, I first found the value of the first constant multiplied by the last constant, which is -8. Then I listed all of the common factors for -8: 1, -1, 2, -2, 4, -4, 8, -8. I tried to use synthetic division and usually, when I get a remainder of 0, I would know that the quotient and divisor are factors of my polynomial but none of my common factors work. When I checked my answer key, it showed me (2x-1)(2x+1) which means x=1/2 and -1/2. So looking for a common factor didn't actually help but I don't know any other ways. Please help me if you know how to solve this problem.
Factoring
- Thread starter Chocolate
- Start date
What do you mean by factor of constant/factor of leading coefficient? I was guessing you were saying -1/8 but that can't be right.... can you explain please?
o and yes, idid learn the remainder test but i keep on forgetting to use it
Thanks for reminding me!
o and yes, idid learn the remainder test but i keep on forgetting to use it
Thanks for reminding me!
It's called the rational root theorem. Info: http://www.mathwords.com/r/rational_root_theorem.htm
In your equation, the leading coefficient is 1. Factors: +/- 1
The constant term is 8. Factors: +/- (1, 2, 4, 8).
So rational roots will come from x = +/- (1, 1/2, 1/4, 1/8).
In your equation, the leading coefficient is 1. Factors: +/- 1
The constant term is 8. Factors: +/- (1, 2, 4, 8).
So rational roots will come from x = +/- (1, 1/2, 1/4, 1/8).
Chocolate said:
Since you state that you haven't seen the Factor Theorem or Remainder Theorem yet, I wonder if you had considered "factoring by grouping".....
Group the first two terms together, and the last two terms together:
8x<SUP>3</SUP> + 4x<SUP>2</SUP> - 2x - 1
Remove a common factor of 4x<SUP>2</SUP> from the first two terms, and a common factor of -1 from the last two terms:
4x<SUP>2</SUP>(2x + 1) -1(2x + 1)
Now, (2x + 1) is a common factor. Remove it:
(2x + 1)(4x<SUP>2</SUP> - 1)
The expression inside the second set of parentheses can be factored further as a difference of two squares.
A few minutes spent checking to see if an expression can be factored by grouping may save you lots of time in the long run.....just my opinion, of course.