Question

Helen and Kiran are trying to find the values of p and q in the equation

x^3+ px^2+32x+q=0. they know it has a single root of x=2 and a repeated root elsewhere. solve to find the two sets of values for p and q

Answer 1

To find the values of p and q in the equation x^3+ px^2+32x+q=0, given a single root of x=2 and a repeated root elsewhere, we can use the fact that the quotient of the polynomial divided by the single root will have no remainder. Setting this quotient equal to zero, we can solve for p and q. The values of p and q are 0 and -44, respectively.

Step-by-step explanation:

The question is asking to find the values of p and q in the equation x^3+ px^2+32x+q=0 given that it has a single root of x=2 and a repeated root elsewhere. To solve this, we can use the fact that if a polynomial has a repeated root, then it must be a factor of the polynomial and the polynomial divided by that root will have no remainder.

Since the given equation has a single root of x=2, we can divide the polynomial by (x-2). Using synthetic division, we get a quotient of x^2 + (p+2)x + (32-2p+q) without any remainder. Since this quotient has a repeated root, it will also be a factor of the quotient. Therefore, we can set this quotient equal to zero and solve for p and q.

x^2 + (p+2)x + (32-2p+q) = 0

Substituting the root x=2, we get:

2^2 + (p+2)(2) + (32-2p+q) = 0

Simplifying, we get:

4 +2p + 4 + 32 -2p + q = 0

Combining like terms, we get:

2p + 40 -2p + q = -4

Simplifying, we get:

q + 40 = -4

Subtracting 40 from both sides, we get:

q = -44

Therefore, the values of p and q are:

p = 0

q = -44

Answer 2

(x - 2)(x - a)²

= (x - 2)(x² - 2ax + a²)

= x³ - 2ax² + a²x - 2x² + 4ax - 2a²

= x³ - 2(a + 1)x² + a(a + 4)x - 2a²

a(a + 4) = 32

a² + 4a - 32 = 0

(a + 8)(a - 4) = 0

a = -8, 4

p = -2(a + 1)

a = -8--->p = -2(-8 + 1) = -2(-7) = 14

a = 4---->p = -2(4 + 1) = -2(5) = -10

q = -2a²

a = -8--->q = -2(-8)² = -2(64) = -128

a = 4---->q = -2(4²) = -2(16) = -32

(p, q) = {(14, -128), (-10, -32)}