SOLUTION: CAN YOU HELP ME. I REALLY DONT UNDERSTAND. I HAVE TO VERIFY THE GIVEN FACTORS OF THE FUNCTION, FIND THE REMAINING FACTORS OF f, USE MY RESULTS TO WRITE THE COMPLETE FACTORIZATION O

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: CAN YOU HELP ME. I REALLY DONT UNDERSTAND. I HAVE TO VERIFY THE GIVEN FACTORS OF THE FUNCTION, FIND THE REMAINING FACTORS OF f, USE MY RESULTS TO WRITE THE COMPLETE FACTORIZATION O      Log On


   

Question 118563: CAN YOU HELP ME. I REALLY DONT UNDERSTAND. I HAVE TO VERIFY THE GIVEN FACTORS OF THE FUNCTION, FIND THE REMAINING FACTORS OF f, USE MY RESULTS TO WRITE THE COMPLETE FACTORIZATION OF f, LIST ALL REAL 0'S OF f, AND CONFRIM MY RESULTS BY USING A GRAPHING ULTILITY TO GRAPH THE FUNCTION.
f(x)=2x(to the 3rd power)+x(to the 2nd power)-5x+2 Factors
(x+2) (x-1)
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
You need to use long division or synthetic division to verify that the given factors are indeed factors of the function. I'll try to walk you through the steps for one of them:

%282x%5E3%2Bx%5E2-5x%2B2%29%2F%28x%2B2%29

The first term of the denominator, x, goes into the first term of the numerator, 2x%5E3, 2x%5E2 times, so that becomes the first term of your quotient. 2x%5E2%2A%28x%2B2%29=2x%5E3%2B4x%5E2 so subtract these two terms from the first two terms of the numerator expression yielding -3x%5E2. Bring down the next numerator term and divide the resulting binomial, -3x%5E2-5x, by x%2B2. The first term of the divisor, x, goes into the first term -3x times, so that becomes the second term of your quotient. -3x%2A%28x%2B2%29=3x%5E2-6x. Subtract these two terms from the (-3x%5E2-5x dividend part resulting in x. Bring down the %2B2 resulting in x%2B2. Divide the x%2B2 by x%2B2 resulting in 1, which becomes the third term of your quotient. The quotient is now 2x%5E2-3x%2B1 and there is no remainder. Since there is no remainder, you have verified that x%2B2 is a factor of %282x%5E3%2Bx%5E2-5x%2B2%29.

Repeat the process using the quotient you just derived, 2x%5E2-3x%2B1, and dividing that by the other given factor, x-1. Since the process is the same as described above, I'll leave the details to you, but you should get a quotient of 2x-1 with no remainder. That tells you that x-1 is a factor and that the third factor is 2x-1.

Now you can say that f%28x%29=2x%5E3%2Bx%5E2-5x%2B2=%28x%2B2%29%28x-1%29%282x-1%29. The zeros of f are those values of x for which f%28x%29+=+0.

f%28x%29+=+0 if and only if x%2B2=0 or x-1=0 or 2x-1=0.

Solving each of these equations yields:

x%5B1%5D=-2
x%5B2%5D=1
x%5B3%5D=1%2F2

graph%28600%2C600%2C-8%2C8%2C-8%2C8%2C2x%5E3%2Bx%5E2-5x%2B2%29

Notice that the graph crosses the x-axis at the three zeros of f.

Notice that this graph has a peak and a valley. It's beyond the scope of this discussion why, but the values of x at the top of the peak and the bottom of the valley are given by solving 6x%5E2%2B2x-5=0, yielding %28-1%2B-sqrt%2831%29%29%2F6, very roughly speaking -1 and 3/4, as you can see on the graph. The y values for these points would obviously be the original function evaluated at these two x values.

Hope that helps.
John