GUIDED PRACTICE

1. ANSWER 4 for Example 1 GUIDED PRACTICE 1. How many solutions does the equation x4 + 5x2– 36 = 0 have?

2. ANSWER 3 for Example 1 GUIDED PRACTICE 2. How many zeros does the function f (x) = x3 + 7x2 + 8x – 16 have?

3. Find the rational zero of f. because f is a polynomial function degree 3, it has 3 zero. The possible rational zeros are 1 , 3, using synthetic division, you can determine that 3 is a zero reputed twice and –3 is also a zero STEP 1 STEP 2 Writef (x) in factored form + + – – for Example 2 GUIDED PRACTICE Find all zeros of the polynomial function. 3. f (x) = x3 + 7x2 + 15x + 9 SOLUTION Formula are (x +1)2 (x +3) f(x) = (x +1) (x +3)2 The zeros of f are – 1 and – 3

4. Find the rational zero of f. because f is a polynomial function degree 5, it has 5 zero. The possible rational zeros are 1 , 2, 3 and Using synthetic division, you can determine that 1 is a zero reputed twice and –3 is also a zero STEP 1 6. STEP 2 Writef (x) in factored formdividing f(x)by its known factor (x – 1),(x – 1)and (x+2) given a qualitiesx2 – 2x +3 therefore + + + + – – – – for Example 2 GUIDED PRACTICE 4. f (x) = x5– 2x4 + 8x2– 13x + 6 SOLUTION f (x) = (x – 1)2 (x+2) (x2 – 2x + 3)

5. STEP 3 Find the complex zero of f. use the quadratic formula to factor the trinomial into linear factor f (x) = (x –1)2 (x + 2) [x – (1 + i 2) [ (x – (1 – i 2)] Zeros of f are 1, 1, – 2, 1 + i 2 , and 1 – i 2 for Example 2 GUIDED PRACTICE

6. GUIDED PRACTICE for Example 3 Write a polynomial functionf of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros. 5. – 1, 2, 4 Use the three zeros and the factor theorem to write f(x) as a product of three factors. SOLUTION f (x) = (x + 1) (x – 2) ( x – 4) Write f (x) in factored form. = (x + 1) (x2– 4x – 2x + 8) Multiply. = (x + 1) (x2– 6x + 8) Combine like terms. = x3 – 6x2 + 8x+ x2– 6x + 8 Multiply. = x3– 5x2 + 2x + 8 Combine like terms.

7. 6. 4, 1 + √5 f (x) = (x – 4) [ x – (1 + √ 5 ) ] [ x – (1 – √ 5 ) ] Because the coefficients are rational and 1 + 5 is a zero, 1 – 5 must also be a zero by the irrational conjugates theorem. Use the three zeros and the factor theorem to write f(x) as a product of three factors = (x – 4) [ (x – 1) – √ 5 ] [ (x – 1) +√ 5 ] = (x – 4)[(x – 1)2 – ( 5)2] GUIDED PRACTICE for Example 3 SOLUTION Write f (x) in factored form. Regroup terms. Multiply. = (x – 4)[(x2 – 2x + 1) – 5] Expand binomial.

8. GUIDED PRACTICE for Example 3 = (x – 4)(x2 – 2x – 4) Simplify. = x3 – 2x2 – 4x – 4x2 + 8x + 16 Multiply. = x3 – 6x2 + 4x +16 Combine like terms.

9. √6 7. 2, 2i, 4 – Because the coefficients are rational and 2i is a zero, –2i must also be a zero by the complex conjugates theorem. 4 + 6 is also a zero by the irrational conjugate theorem. Use the five zeros and the factor theorem to write f(x) as a product of five factors. f (x) = (x–2) (x +2i)(x-2i)[(x –(4–√6 )][x –(4+√6) ] = (x – 2) [ (x2–(2i)2][x2–4)+√6][(x– 4) – √6 ] = (x – 2)[(x2+ 4)[(x– 4)2 – ( 6 )2] GUIDED PRACTICE for Example 3 SOLUTION Write f (x) in factored form. Regroup terms. Multiply. = (x – 2)(x2 + 4)(x2– 8x+16 – 6) Expand binomial.

10. GUIDED PRACTICE for Example 3 = (x – 2)(x2 + 4)(x2– 8x + 10) Simplify. = (x–2) (x4– 8x2 +10x2 +4x2 –3x +40) Multiply. = (x–2) (x4 – 8x3+14x2 –32x + 40) Combine like terms. = x5– 8x4 +14x3 –32x2 +40x– 2x4 +16x3 –28x2 + 64x – 80 Multiply. = x5–10x4 + 30x3 – 60x2 +10x– 80 Combine like terms.

11. GUIDED PRACTICE for Example 3 8. 3, 3 – i Because the coefficients are rational and 3 –i is a zero, 3 + i must also be a zero by the complex conjugates theorem. Use the three zeros and the factor theorem to write f(x) as a product of three factors SOLUTION = f(x) =(x – 3)[x – (3 – i)][x –(3 + i)] Write f (x) in factored form. = (x–3)[(x– 3)+i ][(x– 3) – i] Regroup terms. = (x–3)[(x– 3)2 –i2)] Multiply. = (x– 3)[(x – 3)+ i][(x–3) –i]

12. GUIDED PRACTICE for Example 3 = (x – 3)[(x – 3)2 – i2]=(x –3)(x2 – 6x + 9) Simplify. = (x–3)(x2 – 6x + 9) = x3–6x2 + 9x– 3x2 +18x – 27 Multiply. Combine like terms. = x3 – 9x2 + 27x –27