Best answer:
7.
p(x) = a (x − 2) (x + 2) (x − b) (x − c)
This polynomial has roots: 2, −2, b, c and leading coefficient a.
Since we have 3 unknown quantities, we need 3 conditions to solve.
So we cannot find a unique polynomial where p(0) = 4, p(1) = −3
p(0) = a(−2)(2)(−b)(−c) = 4
−4abc = 4
abc = −1
p(−3) =...
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Best answer:
7.
p(x) = a (x − 2) (x + 2) (x − b) (x − c)
This polynomial has roots: 2, −2, b, c and leading coefficient a.
Since we have 3 unknown quantities, we need 3 conditions to solve.
So we cannot find a unique polynomial where p(0) = 4, p(1) = −3
p(0) = a(−2)(2)(−b)(−c) = 4
−4abc = 4
abc = −1
p(−3) = a(−1)(3)(1−b)(1−c) = −3
−3a(1−b)(1−c) = −3
−a(1−b)(1−c) = −1
abc = −a(1−b)(1−c)
bc = −(1−b)(1−c)
bc = −1 + b + c − bc
c = (b−1)/(2b−1)
abc = −1
ab(b−1)/(2b−1) = −1
ab² − ab = −2b + 1
ab² + (2−a)b − 1 = 0
b = ((a−2) ± √((2−a)²+4a)) / (2a), where a > 0 or a ≤ −4
b = (a − 2 ± √(a²+4)) / (2a)
c = (a − 2 ∓ √(a²+4)) / (2a)
p(x) = 0 ---> x = ± 2, (a − 2 ± √(a²+4)) / (2a)
If leading coefficient a = 1, then we get:
p(x) = (x − 2) (x + 2) (x − (−1+√5)/2) (x − (−1−√5)/2)
p(x) = 0 ----> x = ± 2, (−1±√5)/2
11.
p(x) = x³ + ax² + bx − 18
(x + 2) is a factor of p(x) iff p(−2) = 0
−8 + 4a − 2b − 18 = 0
4a − 2b = 26
2a − b = 13
b = 2a − 13
−24 is remainder when p(x) is divided by (x−1) iff p(1) = −24
1 + a + b − 18 = −24
a + b = −7
a + (2a − 13) = −7
3a = 6
a = 2
b = 2(2)−13 = −9
p(x) = x³ + 2x² − 9x − 18
12.
You have unmatched parentheses. Did you mean: (x+2)(Q(x)+3) or (x+2)Q(x)+3 ?
x³ − 2x² + a = (x+2) (Q(x) + 3)
When x = −2
−8 −8 + a = (−2+2) (Q(−2) + 3)
−16 + a = 0
a = 16
x³ − 2x² + a = (x+2) Q(x) + 3
When x = −2
−8 −8 + a = (−2+2) Q(−2) + 3
−16 + a = 0 + 3
a = 19