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February 1, 2015

February 1, 2015

Posted by **Confused** on Thursday, October 13, 2011 at 6:09pm.

Determine the roots algebraically by factoring.

a) x^3-8x^2-3x+90=0

- Grade 12- Advanced Funtions -
**MathMate**, Thursday, October 13, 2011 at 9:06pmx^3-8x^2-3x+90=0

The leading coefficient is 1, so that makes the factorization simpler by having less combinations to check.

The possible rational factors (if any) are the factors of 90. Since there are variations of signs in the expression, the factors could be + or -.

In fact, the Descartes rule of signs says that there are (maximum) 2 positive roots, and 1 negative root.

Since it is a cubic, there is at least one real root.

Factorization:

Step 1:

Check the coefficients to see if there are ways they cancel out each other. If yes, x+1 or x-1 could be a candidate.

(not in this case).

1-8-3+90 (does not work for x=±1)

Try multiplying the first term by 2^3, second term by 2^2, third term by 2 and repeat above check.

8-32-6+90 (does not work for x=±2)

Try the next factor 3 and repeat above:

27-72-9+90 (does not work for x=3)

They don't add up to zero, but note that -27-72+9+90 would work.

So try x=-3 by changing the sign of coefficients of odd powers, namely the first and third, which gives exactly

-27-72+9+90=0

So x=-3 is a zero.

Use synthetic division to reduce the expression to a quadratic:

x²-11x+30

which I'm sure you can factorize to complete the answer.

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