Given that, x^2-4x+4 is a root of
x^3+ax^2+bx-4. Find a and b
please show workings
#thanks
To find the values of a and b, we need to use the fact that x^2-4x+4 is a root of the equation x^3 + ax^2 + bx - 4.
A root of an equation is a value of x that makes the equation equal to zero. So, if x^2-4x+4 is a root of the equation, then when we substitute x^2-4x+4 into the equation, it should equal zero.
Let's substitute x^2-4x+4 into the equation x^3 + ax^2 + bx - 4:
(x^3 + ax^2 + bx - 4) = 0
Now, substitute x^2-4x+4 for x in the equation:
((x^2-4x+4)^3 + a(x^2-4x+4)^2 + b(x^2-4x+4) - 4) = 0
Expand the equation and simplify:
(x^6 - 12x^5 + 56x^4 - 112x^3 + (80+2a)x^2 + (-32a+4b+56)x + (16a-4b-4)) = 0
Since x^2 - 4x + 4 is a root, we know that substituting it into the equation should equal zero:
((x^2-4x+4)^3 + a(x^2-4x+4)^2 + b(x^2-4x+4) - 4) = 0
Plugging in x^2-4x+4 into the expanded equation, we get:
((x^2-4x+4)^3 - 12(x^2-4x+4)^2 + 56(x^2-4x+4)^4 + (-112(x^2-4x+4))^3 + (80+2a)(x^2-4x+4)^2 + (-32a+4b+56)(x^2-4x+4) + (16a-4b-4)) = 0
Now, we have a polynomial equation in terms of x, and we can equate the coefficients of x to find a and b.
The coefficient of x^2 in the expanded equation is: 56 + (80+2a) + 2a = 56 + 80 + 2a + 2a = 136 + 4a.
Therefore, a = -34.
The coefficient of x in the expanded equation is: (-32a+4b+56) = (-32*(-34) + 4b + 56) = 1088 + 4b.
Therefore, b = -272.
Hence, the values of a and b are a = -34 and b = -272.