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DiKsa [7]
3 years ago
9

1. Make sure your answers and work is SHOWN please and steps are in order.

Mathematics
1 answer:
JulijaS [17]3 years ago
4 0

Been a while since I've done synthetic division.

1a. Let's assume that's supposed to be an equals sign

p(x) = 2x⁴ - 3x³ - 6x² + 5x + 6

possible rational roots have factors of 6 in the numerator and of 2 in the denominator.  We'll only worry about negative numerators.

Factors of six: 1,2,3,6, and we don't forget -1,-2,-3,-6

Factors of 2: 1,2

Possible rational roots:

(dividing by 1:) 1,-1,2,-2,3,-3,6,-6

(dividing by 2:) 1/2, -1/2  (2/2=1 is a duplicate, don't have to repeat it), 3/2, -3/2

Possible rational roots: 1,-1,2,-2,3,-3,6,-6, 1/2, -1/2, 3/2, -3/2

Synthetic division, trying x=1,

   1 | 2  -3  -6  5  6

             2  -1  -7  -2

       2    -1  -7  -2  4

Got a remainder of 4, so 1 isn't a root;

Trying x=-1  

-1 | 2  -3  -6  5  6

         -2   5  1  -6

     2  -5  -1  6  0

Zero remainder, found a root, x=-1.  This division says

(2x⁴ - 3x³ - 6x² + 5x + 6) / (x + 1) = 2x³ - 5x² - x + 6

Same set of rational roots on the cubic, we continue with x=2

2 | 2 -5 -1 6

         4 -2 -6

    2  -1 -3 0

Another zero remainder, x=2 is a root.  We're left with 2x² - x - 3 = 0, which factors as

(2x - 3)(x + 1) = 0

That's a second factor of x+1.  Our final factorization is

2x⁴ - 3x³ - 6x² + 5x + 6 = (x + 1)²(x-2)(2x - 3)

Fourth degree with a positive leading coefficient so goes to +infinity at both ends.  Double zero at x=-1, so it's tangent there, just touching the x axis, then down through x=3/2 and up through x=2.  

I'll leave the actually sketching and the other two polynomials to you -- that took some time.

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