The quadratic function 𝑓(𝑥) which is exactly divisible by 2𝑥 + 1 and has remainder -6 and -5

Question

when divided by 𝑥 − 1 and 𝑥 − 2 respectively. Determine g(x)=(𝑝𝑥 + 𝑞)2𝑓(𝑥) where p and q
are constants and equal. Given that on division by (𝑥 − 2)2,𝑡ℎ𝑒 𝑟𝑒𝑚𝑎𝑛𝑖𝑑𝑒𝑟 − 39 − 3𝑥

Answer 1

We can start by using the factor theorem to find the quadratic factor of f(x) that is exactly divisible by 2x+1. If 2x+1 is a factor, then we know that f(-1/2) = 0. So we have:

f(-1/2) = a(-1/2 - 1)(-1/2 - 2) = a(3/2)(5/2) = 15a/4 = 0

This implies that a = 0, so 2x+1 is not a factor of f(x).

Next, we can use the remaind er theorem to find the quadratic factor of f(x) that has remainder -6 when divided by x-1 and remainder -5 when divided by x-2. We have:

f(1) = a(1-2)(1-3) + c = 2a + c = -6
f(2) = a(2-1)(2-3) + c = -a + c = -5

Solving these equations simultaneously, we get a=-11 and c=16. So the quadratic factor of f(x) is:

f(x) = -11(x-1)(x-2) + 16

Now we can find g(x)=(px+q)^2f(x). Since the quadratic factor of f(x) has a leading coefficient of -11, we want the leading coefficient of g(x) to also be -11. This means that p^2 = -11, so we can take p = i*sqrt(11), where i is the imaginary unit.

To determine q, we'll use the fact that g(x) has a double root at x=2. This means that g(x) is divisible by (x-2)^2, so we can write:

g(x) = (px+q)^2f(x) = (px+q)^2[-11(x-1)(x-2) + 16]
= -11(px+q)^2(x-1)(x-2) + 16(px+q)^2

Taking the derivative of this expression with respect to x, we get:

g'(x) = -22(px+q)(x-1)(x-2) + 32px(px+q)

At x=2, we have:

g'(2) = -22(i*sqrt(11)*2+q)(2-1)(2-2) + 32i*sqrt(11)*2(i*sqrt(11)*2+q) = 0

Simplifying this equation, we get:

-44i*sqrt(11) + 64q = 0

So q = 11i*sqrt(11)/16.

Finally, we can write:

g(x) = (i*sqrt(11)x + 11i*sqrt(11)/16)^2[-11(x-1)(x-2) + 16]
= -11(11x^2/16 - 11ixsqrt(11)x/8 - 121/256)(x-1)(x-2) + 16(11x^2/16 + 11ixsqrt(11)x/8 + 121/256)
= -121(x-2)^2 - 11ixsqrt(11)(x-2)^2 + 176

Now we can use the given information to determine the remainder when g(x) is divided by (x-2)^2:

-121(x-2)^2 - 11ixsqrt(11)(x-2)^2 + 176 = ax + b(x-2)^2 + c(x-2)^3 + 39 + 3x

Expanding both sides and comparing coefficients, we get:

a = -3
b = 0
c = -11/3

So the remainder is:

-121(x-2)^2 - 11ixsqrt(11)(x-2)^2 + 176 = -3x - 11/3(x-2)^3 + 39 + 3x
-121(x-2)^2 - 11ixsqrt(11)(x-2)^2 + 176 = -11/3(x-2)^3 + 39

Solving for ixsqrt(11), we get:

ixsqrt(11) = (-11/3(x-2)^3 + 121(x-2)^2 + 176 - 39)/(x-2)^2
ixsqrt(11) = (-11/3(x-2) + 121) + 97/(x-2)^2

So the expression for g(x) simplifies to:

g(x) = -(11x^2/16 - 11ixsqrt(11)x/8 - 121/256)(x-1)(x-2) + 16(11x^2/16 + 11ixsqrt(11)x/8 + 121/256)
= -11x^2 + 88ixsqrt(11)x + 704/3

Therefore, p= i*sqrt(11) and q = 11i*sqrt(11)/16, and g(x) = -11x^2 + 88ixsqrt(11)x + 704/3.