Find all the zeroes of: f(x)=2x^5+5x^4+4x^3+17x^2-16x-12 show the p/q set, the descartes sign chart, and your synthetic divison table. Then sketch a graph of your results

Answer 1

f(x) = 2x⁵ + 5x⁴ + 4x³ + 17x² - 16x - 12

The Rational Roots Theorem say that IF there are rational roots (zeros), they will be plus/minus factors of 12 (the constant term) divided by the factors of 2 (the leading coefficient) so p = ± 12, 6, 4, 3, 2, 1 and q = ± 2, 1 so the p/q set is ± 12, 6, 4, 3, 2, 3/2, 1, 1/2

Descartes' Rule of Signs:

Sign changes (positive): one sign change = 1 positive real zero

Sign changes (negative): -2x⁵ + 5x⁴ - 4x³ + 17x² + 16x - 12 = 4 sign changes so either 4, 2, or 0 negative real zeros.

Chart:

Pos Neg Imag Total

1 4 0 5

1 2 2 5

1 0 4 5

The synthetic division table would consist of and zeros we find as well as values of the function for the "unsuccessful" tries. I'm unsure how you were instructed to build the table so I'll just do it the way I would. One "unsuccessful" try is x = 0, which is easy to determine since it is just the constant term so f(0) = -12

For Synthetic Division, I like to start easy (like 1 or -1) so I'll try 1:

1 | 2 5 4 17 -16 -12

2 7 11 28 12

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2 7 11 28 12 0

It works, so x = 1 is the ONE positive real zero

Now -1:

-1 | 2 5 4 17 -16 -12

-2 -3 -1 -16 32

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2 3 1 16 -32 20

So we are unsuccessful but f(-1) = 20

However, since f(0) = -12 and f(-1) = 20 there is a zero between them and since -1/2 is a possible choice, let's try that:

-1/2 | 2 5 4 17 -16 -12

-1 -2 -1 -8 12

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2 4 2 16 -24 0

So success again x = -1/2 is a zero.

So let's try another value smaller than -1. I'll try -3/2:

-3/2 | 2 5 4 17 -16 -12

-3 -3 -1.5 -23.25 58.875

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2 2 1 15.5 -39.25 46.875

So we are unsuccessful but f(-3/2) = 46.875

So let's try another value smaller than -3/2. I'll try -2:

-2 | 2 5 4 17 -16 -12

-4 -2 -4 -26 84

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2 1 2 13 -42 72

So we are unsuccessful but f(-2) = 72

So let's try another value smaller than -2. I'll try -3:

-3 | 2 5 4 17 -16 -12

-6 3 -21 12 12

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2 -1 7 -4 -4 0

So success again x = -3 is a zero

So let's try another value smaller than -3. I'll try -4:

-4 | 2 5 4 17 -16 -12

-8 12 -64 188 -688

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2 -3 16 -47 172 -700

We are unsuccessful again but f(-4) = -700 however, even more important is that x=-4is a lower bound because we get the pattern of + / - / + / - / + / - in the coefficients of the quotient. So not only is -4 not a zero, but there are no zeros less than that.

So the table would consist of all these results.

So we have zeros of x = 1, x = -1/2, and x = -3. The resulting quadratic can be found dividing these in a cascading type manner:

1 | 2 5 4 17 -16 -12

2 7 11 28 12

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-1/2 | 2 7 11 28 12

-1 -3 -4 -12

-----------------------------------

-3 | 2 6 8 24

-6 0 -24

---------------------------

2 0 8

So (2x² + 8) is the final factor

To find the final complex (imaginary) zeros set 2x² + 8 equal to zero and solve:

2x² + 8 = 0

2x² = -8

x² = -4

x = ±√-4 = ±2i

All 5 zeros are then x = 1, x = -1/2, x = -3, x = 2i, and x = -2i

A sketch of the graph could be: