Question

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

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

[AI-generated answer — Bot]

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.

Answer 2

[AI-generated answer — Step-by-Step Bot]

To find the quadratic function 𝑓(𝑥) exactly divisible by 2𝑥 + 1, we need to use the factor theorem. According to the factor theorem, if 𝑥 − 𝑎 is a factor of a polynomial 𝑓(𝑥), then 𝑓(𝑎) = 0.

In this case, 2𝑥 + 1 is a factor, so we can set it equal to zero and solve for 𝑥:

2𝑥 + 1 = 0
2𝑥 = -1
𝑥 = -1/2

So, 𝑥 = -1/2 is a root of the quadratic function 𝑓(𝑥).

Now, let's write 𝑓(𝑥) in the form of (𝑥 − 𝑎)(𝑥 − 𝑏), where 𝑎 and 𝑏 are the other roots of 𝑓(𝑥) (not including 𝑥 = -1/2).

Since 𝑥 − 1 and 𝑥 − 2 are factors of 𝑓(𝑥), we have:

𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)𝑔(𝑥)

where 𝑔(𝑥) is another quadratic function.

Next, we can find the value of 𝑓(𝑥) when divided by 𝑥 − 1 and 𝑥 − 2 respectively.

When divided by 𝑥 − 1, the remainder is -6. This means 𝑓(1) = -6:

𝑓(1) = (1 − 1)(1 − 2)𝑔(1) = 0𝑔(1) = 0

Since 𝑔(1) cannot be zero, we can conclude that (1 − 1)(1 − 2) must equal -6:

0 = (1 − 1)(1 − 2)
0 = (0)(-1)
0 = 0

This is true, so 𝑔(1) can take any value. Let's assume 𝑔(1) = 𝑞, where 𝑞 is a constant.

Now, when divided by 𝑥 − 2, the remainder is -5. This means 𝑓(2) = -5:

𝑓(2) = (2 − 1)(2 − 2)𝑔(2) = 𝑔(2) = -5

So, 𝑔(2) = -5. Substituting 𝑥 = 2 into 𝑔(𝑥) = 𝑞, we have:

𝑞 = -5

Now, let's find the expression for 𝑔(𝑥):

𝑔(𝑥) = -5

Finally, let's find 𝑔(𝑥) in terms of 𝑝 and 𝑞:

𝑔(𝑥) = -5 = 𝑞
𝑔(𝑥) = 𝑞 = -5

Since 𝑝 and 𝑞 are constants and equal, 𝑝𝑞 = 𝑞².

So, 𝑔(𝑥) = 𝑞²

Therefore, g(𝑥) = (𝑝𝑥 + 𝑞)²𝑓(𝑥) = (𝑝𝑥 - 5)²(𝑥 − 1)(𝑥 − 2)