Substitution method: which linear expression is NOT a factor
- Thread starter fencer817
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Sadly, there is nothing to solve. Did you mean:
3x^4 - 10x^3 - 29x^2 + 16x + 20 = 0?
If there are Rational Number solutions, they are factors of 20 divided by factors of 3. So,
20, 20/3
10, 10/3
5, 5/3
4, 4/3
2,2/3
1,1/3
Positive or negative. There! The possibilities for Rational Number solutions are whittled down to only 24.
There is a sneaky trick for this sort of thing. It's called Horner's Method. It requires LOTS of parentheses to describe, but only a litte effort to demonstrate.
Let's try x = 1, jut for fun. Watch the coefficients fly...
3x^4 - 10x^3 - 29x^2 + 16x + 20 = 0
3*1 = 3
(3-10)*1 = -7
(-7-29)*1 = -36
(-36+16)*1 = -20
(-20+20) = 0 <== No "*1" on the last one.
Cool!!! We found a solution. Guess what else we found? The factorization that goes with that solution. Check out the values on the right of the equal signs.
3x^4 - 10x^3 - 29x^2 + 16x + 20 = (x-1)(3x^3 - 7x^2 - 36x - 20) = 0
Since the new cubic polynomial has the same leading coefficient and constant term, start over with the same list.
Note: In case someone is wondering, Horner's Method should look awfully familiar to Synthetic Division users.
3x^4 - 10x^3 - 29x^2 + 16x + 20 = 0?
If there are Rational Number solutions, they are factors of 20 divided by factors of 3. So,
20, 20/3
10, 10/3
5, 5/3
4, 4/3
2,2/3
1,1/3
Positive or negative. There! The possibilities for Rational Number solutions are whittled down to only 24.
There is a sneaky trick for this sort of thing. It's called Horner's Method. It requires LOTS of parentheses to describe, but only a litte effort to demonstrate.
Let's try x = 1, jut for fun. Watch the coefficients fly...
3x^4 - 10x^3 - 29x^2 + 16x + 20 = 0
3*1 = 3
(3-10)*1 = -7
(-7-29)*1 = -36
(-36+16)*1 = -20
(-20+20) = 0 <== No "*1" on the last one.
Cool!!! We found a solution. Guess what else we found? The factorization that goes with that solution. Check out the values on the right of the equal signs.
3x^4 - 10x^3 - 29x^2 + 16x + 20 = (x-1)(3x^3 - 7x^2 - 36x - 20) = 0
Since the new cubic polynomial has the same leading coefficient and constant term, start over with the same list.
Note: In case someone is wondering, Horner's Method should look awfully familiar to Synthetic Division users.
Thanks that helps and to answer your question. No, sadly she didn't put an =0 on the end. The question stated: Use substitution to determine which of the given linear expressions is not a factor of 3x^4 - 10x^3 - 29x^2 + 16x + 20. Then she listed the answers. She left me hanging to solve it and i didn't even know what i was solving.
