Question 1 :
Find a number b such that 3 is a zero of the polynomial p defined by p(x) = 1 − 4x + bx2 + 2x3
Solution :
p(x) = 1 − 4x + bx2 + 2x3
If x = 3, then p(3) = 0
p(3) = 1 − 4(3) + b(3)2 + 2(3)3
0 = 1 - 12 + 9b + 2(27)
0 = -11 + 9b + 54
0 = 43 + 9b
9b = -43
b = -43/9
Question 2 :
Find a number c such that −2 is a zero of the polynomial p defined by p(x) = 5 − 3x + 4x2 + cx3
Solution :
p(x) = 5 − 3x + 4x2 + cx3
If x = -2, then p(-2) = 0
p(-2) = 5 − 3(-2) + 4(-2)2 + c(-2)3
0 = 5 + 6 + 4(4) + c(-8)
0 = 27 - 8c
8c = 27
c = 27/8
Question 3 :
Find all choices of b, c, and d such that 1 and 4 are the only zeros of the polynomial p defined by p(x) = x3 + bx2 + cx + d.
Solution :
p(x) = x3 + bx2 + cx + d
The zeroes of the cubic polynomials are 1 and 4
x = 1, x = 4
By writing them as factors, we get (x - 1) and (x - 4). But we need to form a cubic equation. So, let us write (x - 1) twice or (x - 4) twice.
p(x) = (x - 1)2 (x - 4)
= (x2 - 2x + 1) (x - 4)
= x3 - 4x2 - 2x2 + 8x + x - 4
p(x) = x3 - 6x2 + 9x - 4
p(x) = x3 + bx2 + cx + d
b = -6, c = 9 and d = -4
(or)
p(x) = (x - 4)2 (x - 1)
= (x2 - 8x + 16) (x - 1)
= x3 - 8x2 + 16x - x2 + 8x - 16
p(x) = = x3 - 8x2 - x2 + 16x + 8x - 16
p(x) = = x3 - 9x2 + 24x - 16
p(x) = x3 + bx2 + cx + d
b = -9, c = 24 and d = -16
Question 4 :
Find all choices of b, c, and d such that −3 and 2 are the only zeros of the polynomial p defined by p(x) = x3 + bx2 + cx + d.
Solution :
p(x) = x3 + bx2 + cx + d
The zeroes of the cubic polynomials are -3 and 2
x = -3, x = 2
By writing them as factors, we get (x + 3) and (x - 2). But we need to form a cubic equation. So, let us write (x + 3) twice or (x - 2) twice.
p(x) = (x + 3)2 (x - 2)
= (x2 + 6x + 9) (x - 2)
= x3 - 2x2 + 6x2 - 12x + 9x - 18
= x3 + 4x2 - 3x - 18
p(x) = x3 + bx2 + cx + d
b = 4, c = -3 and d = -18
p(x) = (x - 2)2 (x + 3)
= (x2 - 4x + 4) (x + 3)
= x3 + 3x2 - 4x2 - 12x + 4x + 12
= x3 - x2 - 8x + 12
p(x) = x3 + bx2 + cx + d
b = -1, c = -8 and d = 12
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