Use the Upper and Lower Bound Tests

Determine an interval in which all real zeros of f(x) = x4 – 4x3 – 11x2 – 4x – 12 must lie. Explain your reasoning using the upper and lower bound tests. Then find all the real zeros.

Use the Upper and Lower Bound Tests

Step 2 Test a lower bound of c = –3 and an upper bound of c = 7.

–3 1 –4 –11 –4 –12–3 21 –30 102

1 –7 10 –34 90 Values alternate signs in the last line, so –3 is a lower bound.

7 1 –4 –11 –4 –127 21 70 462

1 3 10 66 450 Values are all nonnegative in last line, so 7 is an upper bound.

Use the Upper and Lower Bound Tests

Step 3 Use the Rational Zero Theorem.Possible rational zeros: Factors of 12 = ±1, ±2, ±3, ±4 , ±6, ±12 .Because the real zeros are in the interval [–3, 7], you can narrow this list to just –1, –2, –3, 1, 2, 3, 4, and 6. From the graph it appears that only –2 and 6 are reasonable.

Use the Upper and Lower Bound Tests

Begin by testing 6.

Now test –2 in the depressed polynomial.

6 1 –4 –11 –4 –126 12 6 12

1 2 1 2 0

–2 1 2 1 2–2 0 –2

1 0 1 0

Use the Upper and Lower Bound Tests

By the division algorithm, f(x) = (x – 6)(x + 2)(x2 + 1). Notice that the factor x2 + 1 has no real zeros associated with it because x2 + 1 = 0 has no real solutions. So f has two real solutions that are both rational, x = –2 and x = 6. The graph of f(x) = x4 – 4x3 – 11x 2 – 4x – 12 supports this conclusion.

Answer: Upper and lower bounds may vary. Sample answer: [–3, 7]; With synthetic division, the values alternate signs when testing –3, and are all nonnegative when testing 7. So, –3 is a lower bound and 7 is an upper bound. The zeros are –2 and 6.

Determine an interval in which all real zeros of f(x) = 2x4 – 5x3 – 13x2 + 26x – 10 must lie. Then find all the real zeros.

A. [0, 4]; 1, 2

B. [–1, 2]; 1,

C. [–3, 5]; 1,

D. [–2, 1]; 1,

Use Descartes’ Rule of Signs

Describe the possible real zeros of f(x) = x4 – 3x3 – 5x2 + 2x + 7.Examine the variations of sign for f(x) and for f(–x).

f(–x) = (–x)4 – 3(–x)3 – 5(–x)2 + 2(–x) + 7

f(x) = x4 – 3x3 – 5x2 + 2x + 7

= x4 + 3x3 – 5x2 – 2x + 7

Answer: 2 or 0 positive real zeros, 2 or 0 negative real zeros

Use Descartes’ Rule of Signs

The original function f(x) has two variations in sign, while f(–x) also has two variations in sign. By Descartes' Rule of Signs, you know that f(x) has either 2 or 0 positive real zeros and either 2 or 0 negative real zeros.

Describe the possible real zeros of g(x) = –x3 + 8x2 – 7x + 9.

A. 3 or 1 positive real zeros, 1 negative real zero

B. 3 or 1 positive real zeros, 0 negative real zeros

C. 2 or 0 positive real zeros, 0 negative real zeros

D. 2 or 0 positive real zeros, 1 negative real zero

Write a polynomial function of least degree with real coefficients in standard form that has –1, 2, and 2 – i as zeros.Because 2 – i is a zero and the polynomial is to have real coefficients, you know that 2 + i must also be a zero. Using the Linear Factorization Theorem and the zeros –1, 2, 2 – i, and 2 + i, you can write f(x) as follows:f(x) = a[x – (–1)](x – 2)[x – (2 – i)](x – (2 + i)]

Find a Polynomial Function Given Its Zeros

Answer: Sample answer: f(x) = x4 – 5x3 + 7x2 + 3x – 10

Find a Polynomial Function Given Its Zeros

While a can be any nonzero real number, it is simplest to let a = 1. Then write the function in standard form.f(x) = (1)(x + 1)(x – 2)[x – (2 – i)][x – (2 + i)] Let a = 1.

= (x2 – x – 2)(x2 – 4x + 5) Multiply.= x4 – 5x3 + 7x2 + 3x – 10 Multiply.

Therefore, a function of least degree that has –1, 2, and 2 – i as zeros is f(x) = x4 – 5x3 + 7x2 + 3x – 10 or any nonzero multiple of f(x).

Write a polynomial function of least degree with real coefficients in standard form that has –2 (multiplicity 2), 0, and 3i as zeros.

A. f(x) = x5 + 4x4 + 13x3 + 36x2 + 36x

B. f(x) = x5 + 4x4 + 9x3 + 18x

C. f(x) = x3 + 2x2 – 3ix2 – 6xi

D. f(x) = x4 + 4x3 – 5x2 – 36x – 36