If x-2 is a factor of x^2 - bx + b, where b is a constant, what is the value of b?

The value of b that makes x-2 a factor of x² - bx + b is 4.

Polynomial is an expression involving the operations of addition, subtraction, multiplication of variables.

Types of polynomials are quadratic, linear, cubic and so on.

The factor theorem states that; when f(k) = 0, then (x – k) is a factor of f(x).

Given that x-2 is a factor of x² - bx + b, hence:

f(2) = 2² - b(2) + b = 0

4 - 2b + b = 0

b = 4

Therefore the value of b that makes x-2 a factor of x² - bx + b is 4.

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we have that

we know that

If is a factor of

then

for

is equal to zero

so

therefore

the answer is

the value of b is equal to

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube