(a) The leading coefficient of P is 1, so all the rational zeros are integers: They are divisors of the constant term 10. Thus the possible candidates are 1 -5 -5 23 10 1 -4 -9 14 1 -4 -9 14 24 +1, ±2, ±5, ±10 Using synthetic division (see the margin), we find that 1 and 2 are not zeros but that 5 is a zero and that P factors as 1 -5 -5 23 10 -6 -22 12 x* - 5x – 5x2 + 23x + 10 (x – 5)(x³ – 5x – 2) 1 -3 -11 We now try to factor the quotient x- 5x - 2. Its possible zeros are the divisors of -2, namely, 1 -5 -5 23 10 0 - 25 - 10 -5 -2 +1, ±2 Since we already know that 1 and 2 are not zeros of the original polynomial P, we don't need to try them again. Checking the remaining candidates, -1 and -2, we see that -2 is a zero (see the margin), and P factors as -2 1 0-5 -2 x* - 5x - 5x? + 23x + 10 (x - 5)(x³ – 5x - 2) %3D -2 4 2 = (x – 5)(x + 2)(x² – 2x - 1) 1 -2 -1 Now we use the Quadratic Formula to obtain the two remaining zeros of P: 2 + V(-2)² – 4(1)(-1) x = = 1+ V2 50 The zeros of P are 5, -2,1 + V2, and 1 - - V2.

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Chapter3: Polynomial And Rational Functions
Section3.CR: Chapter Review
Problem 9CC
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Find all the real zeros of teh polynomial. Use the Quadratic Formula if necessary given.

P(x) = x4 - 6x3 + 4x2 + 15x + 4

(a) The leading coefficient of P is 1, so all the rational zeros are integers: They are
divisors of the constant term 10. Thus the possible candidates are
1 -5
-5
23
10
1
-4
-9
14
1
-4
-9
14
24
+1, ±2, ±5, ±10
Using synthetic division (see the margin), we find that 1 and 2 are not zeros but
that 5 is a zero and that P factors as
1 -5
-5
23
10
-6
-22
12
x* - 5x – 5x2 + 23x + 10 (x – 5)(x³ – 5x – 2)
1
-3
-11
We now try to factor the quotient x- 5x - 2. Its possible zeros are the divisors
of -2, namely,
1 -5
-5
23
10
0 - 25
- 10
-5
-2
+1, ±2
Since we already know that 1 and 2 are not zeros of the original polynomial P,
we don't need to try them again. Checking the remaining candidates, -1 and -2,
we see that -2 is a zero (see the margin), and P factors as
-2 1
0-5 -2
x* - 5x - 5x? + 23x + 10 (x - 5)(x³ – 5x - 2)
%3D
-2
4 2
= (x – 5)(x + 2)(x² – 2x - 1)
1
-2 -1
Now we use the Quadratic Formula to obtain the two remaining zeros of P:
2 + V(-2)² – 4(1)(-1)
x =
= 1+ V2
50
The zeros of P are 5, -2,1 + V2, and 1 -
- V2.
Transcribed Image Text:(a) The leading coefficient of P is 1, so all the rational zeros are integers: They are divisors of the constant term 10. Thus the possible candidates are 1 -5 -5 23 10 1 -4 -9 14 1 -4 -9 14 24 +1, ±2, ±5, ±10 Using synthetic division (see the margin), we find that 1 and 2 are not zeros but that 5 is a zero and that P factors as 1 -5 -5 23 10 -6 -22 12 x* - 5x – 5x2 + 23x + 10 (x – 5)(x³ – 5x – 2) 1 -3 -11 We now try to factor the quotient x- 5x - 2. Its possible zeros are the divisors of -2, namely, 1 -5 -5 23 10 0 - 25 - 10 -5 -2 +1, ±2 Since we already know that 1 and 2 are not zeros of the original polynomial P, we don't need to try them again. Checking the remaining candidates, -1 and -2, we see that -2 is a zero (see the margin), and P factors as -2 1 0-5 -2 x* - 5x - 5x? + 23x + 10 (x - 5)(x³ – 5x - 2) %3D -2 4 2 = (x – 5)(x + 2)(x² – 2x - 1) 1 -2 -1 Now we use the Quadratic Formula to obtain the two remaining zeros of P: 2 + V(-2)² – 4(1)(-1) x = = 1+ V2 50 The zeros of P are 5, -2,1 + V2, and 1 - - V2.
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