The quadratic function ๐(๐ฅ) which is exactly divisible by 2๐ฅ + 1 and has remainder -6 and -5
when divided by ๐ฅ โ 1 and ๐ฅ โ 2 respectively. Determine g(x)=(๐๐ฅ + ๐)2๐(๐ฅ) where p and q
are constants and equal. Given that on division by (๐ฅ โ 2)2,๐กโ๐ ๐๐๐๐๐๐๐๐๐ โ 39 โ 3๐ฅ
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.
To find the quadratic function ๐(๐ฅ) exactly divisible by 2๐ฅ + 1, we need to use the factor theorem. According to the factor theorem, if ๐ฅ โ ๐ is a factor of a polynomial ๐(๐ฅ), then ๐(๐) = 0.
In this case, 2๐ฅ + 1 is a factor, so we can set it equal to zero and solve for ๐ฅ:
2๐ฅ + 1 = 0
2๐ฅ = -1
๐ฅ = -1/2
So, ๐ฅ = -1/2 is a root of the quadratic function ๐(๐ฅ).
Now, let's write ๐(๐ฅ) in the form of (๐ฅ โ ๐)(๐ฅ โ ๐), where ๐ and ๐ are the other roots of ๐(๐ฅ) (not including ๐ฅ = -1/2).
Since ๐ฅ โ 1 and ๐ฅ โ 2 are factors of ๐(๐ฅ), we have:
๐(๐ฅ) = (๐ฅ โ 1)(๐ฅ โ 2)๐(๐ฅ)
where ๐(๐ฅ) is another quadratic function.
Next, we can find the value of ๐(๐ฅ) when divided by ๐ฅ โ 1 and ๐ฅ โ 2 respectively.
When divided by ๐ฅ โ 1, the remainder is -6. This means ๐(1) = -6:
๐(1) = (1 โ 1)(1 โ 2)๐(1) = 0๐(1) = 0
Since ๐(1) cannot be zero, we can conclude that (1 โ 1)(1 โ 2) must equal -6:
0 = (1 โ 1)(1 โ 2)
0 = (0)(-1)
0 = 0
This is true, so ๐(1) can take any value. Let's assume ๐(1) = ๐, where ๐ is a constant.
Now, when divided by ๐ฅ โ 2, the remainder is -5. This means ๐(2) = -5:
๐(2) = (2 โ 1)(2 โ 2)๐(2) = ๐(2) = -5
So, ๐(2) = -5. Substituting ๐ฅ = 2 into ๐(๐ฅ) = ๐, we have:
๐ = -5
Now, let's find the expression for ๐(๐ฅ):
๐(๐ฅ) = -5
Finally, let's find ๐(๐ฅ) in terms of ๐ and ๐:
๐(๐ฅ) = -5 = ๐
๐(๐ฅ) = ๐ = -5
Since ๐ and ๐ are constants and equal, ๐๐ = ๐ยฒ.
So, ๐(๐ฅ) = ๐ยฒ
Therefore, g(๐ฅ) = (๐๐ฅ + ๐)ยฒ๐(๐ฅ) = (๐๐ฅ - 5)ยฒ(๐ฅ โ 1)(๐ฅ โ 2)