# Consider the polynomial P(x) = (x-a)^2 Q(x) where Q(x) is a polynomial.i. if P(x) = 0 has repeated root (double root) at x=a, show P'(a) = 0ii. Find a and b if the equation x^3-ax^2+bx-12=0...

Consider the polynomial **P(x) = (x-a)^2 Q(x)** where Q(x) is a polynomial.**i.** if P(x) = 0 has repeated root (double root) at x=a, show P'(a) = 0**ii.** Find a and b if the equation x^3-ax^2+bx-12=0 has x=2 as a double root.**iii.** Find all roots of P(x) =0

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i. You need to differentiate with respect to x the polynomial function `P(x)` , such that:

`P'(x) = 2(x - a)*Q(x) + (x - a)^2*Q'(x)`

Replacing a for x in `P'(x) ` yields:

`P'(a) = 2(a - a)*Q(x) + (a - a)^2*Q'(x) => P'(a) = 0 + 0 => P'(a) = 0`

**Hence, testing if `P'(a) = 0` , if `P(a) = 0` yields that the statement holds.**

ii. You need to evaluate a,b in polynomial `x^3-ax^2+bx-12=P(x)` using the information provided by the problem that `x = 2` is a double root.

Since `x = 2` is a double root, hence, `P(2) = 0` and `P'(2) = 0` , such that:

`P(2) = 2^3 - 2^2*a + 2b - 12 => 8 - 4a + 2b - 12 = 0 => - 4a + 2b = 4`

`P'(x) = 3x^2-2ax+b => P'(2) = 3*2^2 - 2*2*a + b = 0`

`12 - 4a + b = 0 => - 4a + b = -12`

You need to solve the following system of equations, such that:

`{(- 4a + 2b = 4),(- 4a + b = -12):} => 2b - b = 4 + 12`

`b = 16 => -4a = -12 - 16 => -4a = -28 => a = 7`

**Hence, evaluating the a,b, under the given conditions, yields **`a = 7, b = 16.`