Verified Answer :

calculista

we have that

\(x^2 - bx + b\)


we know that

If \((x-2)\) is a factor of \(x^2 - bx + b\)

then

for \(x=2\)

\(x^2 - bx + b\) is equal to zero

so

\(2^2 - b*2 + b=0\)

\(4 - 2b + b=0\)

\(b=4\)

therefore


the answer is

the value of b is equal to \(4\)

raphealnwobi

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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