In each of the first six problems, take p(x) as the first polynomial, q(x) as the second polynomial and compute the following.
p(1) q(1); p(0) q(0); p(1) q(−1); p(−1) q(1)
In each of the first six problems, take p(x) as the first polynomial, q(x) as the second polynomial and compute the following.
p(1) q(1); p(0) q(0); p(1) q(−1); p(−1) q(1)
(1)
p(x) = (3x + 3)
q(x) = (3x + 2)
p(x) q(x) = 9x2 + 15x + 6
(i) p(1) q(1) = 9(1)2 + 15(1) + 6
= 9 + 15 + 6
= 30
(ii) p(0) q(0) = 9(0)2 + 15(0) + 6 = 6
(iii) p(1) = 3 × 1 + 3
= 3 + 3
= 6
q(−1) = 3 × (−1) + 2
= −3 + 2
= −1
∴ p(1) q(−1) = 6 × (−1) = −6
(iv) p(−1) = 3 × (−1) + 3
= −3 + 3
= 0
q(1) = 3 × 1 + 2
= 3 + 2
= 5
∴ p(−1) q(1) = 0 × 5 = 0
(2)
p(x) = (3x − 4)
q(x) = (4x − 3)
p(x) q(x) = 12x2 − 25x + 12
(i) p(1) q(1) = 12(1)2 − 25(1) + 12
= 12 − 25 + 12
= 24 − 25
= −1
(ii) p(0) q(0) = 12(0)2 − 25(0) + 12 = 12
(iii) p(1) = 3 × 1 − 4
= 3 − 4
= −1
q(−1) = 4 × (−1) − 3
= −4 − 3
= −7
∴ p(1) q(−1) = −1 × (−7) = 7
(iv) p(−1) = 3 × (−1) − 4
= −3 − 4
= −7
q(1) = 4 × 1 − 3
= 4 − 3
= 1
∴ p(−1) q(1) = −7 × 1 = −7
(3)
p(x) = (2x − 3)
q(x) = (4x2 + 6x + 9)
p(x) q(x) = 8x3 − 27
(i) p(1) q(1) = 8(1)3 − 27
= 8 − 27
= −19
(ii) p(0) q(0) = 8(0)3 − 27 = −27
(iii) p(1) = 2 × 1 − 3
= 2 − 3
= −1
q(−1) = 4(−1)2 + 6(−1) + 9
= 4 − 6 + 9
= 13 − 6
= 7
∴ p(1) q(−1) = −1 × 7 = −7
(iv) p(−1) = 2 × (−1) − 3
= −2 − 3
= −5
q(1) = 4(1)2 + 6(1) + 9
= 4 + 6 + 9
= 19
∴ p(−1) q(1) = −5 × 19 = −95
(4)
p(x) = (2x + 3)
q(x) = (4x2 − 6x + 9)
p(x) q(x) = 8x3 + 27
(i) p(1) q(1) = 8(1)3 + 27 = 8 + 27 = 35
(ii) p(0) q(0) = 8(0)3 + 27 = 27
(iii) p(1) = 2 × 1 + 3
= 2 + 3
= 5
q(−1) = 4(−1)2 − 6(−1) + 9
= 4 + 6 + 9
= 19
∴ p(1) q(−1) = 5 × 19 = 95
(iv) p(−1) = 2 × (−1) + 3
= −2 + 3
= 1
q(1) = 4(1)2 − 6(1) + 9
= 4 − 6 + 9
= 13 − 6
= 7
∴ p(−1) q(1) = 1 × 7 = 7
(5)
p(x) = (x − 1)
q(x) = (x2 + x + 1)
p(x) q(x) = x3 − 1
(i) p(1) q(1) = (1)3 − 1
= 1 − 1
= 0
(ii) p(0) q(0) = (0)3 − 1
= −1
(iii) p(1) = 1 − 1 = 0
q(−1) = (−1)2 + (−1) + 1
= 1 − 1 + 1
= 2 − 1
= 1
∴ p(1) q(−1) = 0 × 1 = 0
(iv) p(−1) = −1 − 1= −2
q(1) = (1)2 + (1) + 1
= 1 + 1 + 1
= 3
∴ p(−1) q(1) = −2 × 3 = −6
(6)
p(x) = (x + 1)
q(x) = (x2 − x + 1)
p(x) q(x) = x3 + 1
(i) p(1) q(1) = (1)3 + 1
= 1 + 1
= 2
(ii) p(0) q(0) = (0)3 + 1
= 0 + 1
= 1
(iii) p(1) = 1 + 1 = 2
q(−1) = (−1)2 − (−1) + 1
= 1 + 1 + 1
= 3
∴ p(1) q(−1) = 2 × 3 = 6
(iv) p(−1) = −1 + 1= 0
q(1) = (1)2 − 1 + 1
= 2 − 1
= 1
∴ p(−1) q(1) = 0 × 1 = 0
(1)
p(x) = (3x + 3)
q(x) = (3x + 2)
p(x) q(x) = 9x2 + 15x + 6
(i) p(1) q(1) = 9(1)2 + 15(1) + 6
= 9 + 15 + 6
= 30
(ii) p(0) q(0) = 9(0)2 + 15(0) + 6 = 6
(iii) p(1) = 3 × 1 + 3
= 3 + 3
= 6
q(−1) = 3 × (−1) + 2
= −3 + 2
= −1
∴ p(1) q(−1) = 6 × (−1) = −6
(iv) p(−1) = 3 × (−1) + 3
= −3 + 3
= 0
q(1) = 3 × 1 + 2
= 3 + 2
= 5
∴ p(−1) q(1) = 0 × 5 = 0
(2)
p(x) = (3x − 4)
q(x) = (4x − 3)
p(x) q(x) = 12x2 − 25x + 12
(i) p(1) q(1) = 12(1)2 − 25(1) + 12
= 12 − 25 + 12
= 24 − 25
= −1
(ii) p(0) q(0) = 12(0)2 − 25(0) + 12 = 12
(iii) p(1) = 3 × 1 − 4
= 3 − 4
= −1
q(−1) = 4 × (−1) − 3
= −4 − 3
= −7
∴ p(1) q(−1) = −1 × (−7) = 7
(iv) p(−1) = 3 × (−1) − 4
= −3 − 4
= −7
q(1) = 4 × 1 − 3
= 4 − 3
= 1
∴ p(−1) q(1) = −7 × 1 = −7
(3)
p(x) = (2x − 3)
q(x) = (4x2 + 6x + 9)
p(x) q(x) = 8x3 − 27
(i) p(1) q(1) = 8(1)3 − 27
= 8 − 27
= −19
(ii) p(0) q(0) = 8(0)3 − 27 = −27
(iii) p(1) = 2 × 1 − 3
= 2 − 3
= −1
q(−1) = 4(−1)2 + 6(−1) + 9
= 4 − 6 + 9
= 13 − 6
= 7
∴ p(1) q(−1) = −1 × 7 = −7
(iv) p(−1) = 2 × (−1) − 3
= −2 − 3
= −5
q(1) = 4(1)2 + 6(1) + 9
= 4 + 6 + 9
= 19
∴ p(−1) q(1) = −5 × 19 = −95
(4)
p(x) = (2x + 3)
q(x) = (4x2 − 6x + 9)
p(x) q(x) = 8x3 + 27
(i) p(1) q(1) = 8(1)3 + 27 = 8 + 27 = 35
(ii) p(0) q(0) = 8(0)3 + 27 = 27
(iii) p(1) = 2 × 1 + 3
= 2 + 3
= 5
q(−1) = 4(−1)2 − 6(−1) + 9
= 4 + 6 + 9
= 19
∴ p(1) q(−1) = 5 × 19 = 95
(iv) p(−1) = 2 × (−1) + 3
= −2 + 3
= 1
q(1) = 4(1)2 − 6(1) + 9
= 4 − 6 + 9
= 13 − 6
= 7
∴ p(−1) q(1) = 1 × 7 = 7
(5)
p(x) = (x − 1)
q(x) = (x2 + x + 1)
p(x) q(x) = x3 − 1
(i) p(1) q(1) = (1)3 − 1
= 1 − 1
= 0
(ii) p(0) q(0) = (0)3 − 1
= −1
(iii) p(1) = 1 − 1 = 0
q(−1) = (−1)2 + (−1) + 1
= 1 − 1 + 1
= 2 − 1
= 1
∴ p(1) q(−1) = 0 × 1 = 0
(iv) p(−1) = −1 − 1= −2
q(1) = (1)2 + (1) + 1
= 1 + 1 + 1
= 3
∴ p(−1) q(1) = −2 × 3 = −6
(6)
p(x) = (x + 1)
q(x) = (x2 − x + 1)
p(x) q(x) = x3 + 1
(i) p(1) q(1) = (1)3 + 1
= 1 + 1
= 2
(ii) p(0) q(0) = (0)3 + 1
= 0 + 1
= 1
(iii) p(1) = 1 + 1 = 2
q(−1) = (−1)2 − (−1) + 1
= 1 + 1 + 1
= 3
∴ p(1) q(−1) = 2 × 3 = 6
(iv) p(−1) = −1 + 1= 0
q(1) = (1)2 − 1 + 1
= 2 − 1
= 1
∴ p(−1) q(1) = 0 × 1 = 0
