contestada

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

Respuesta :

calculista

we have that

[tex] x^2 - bx + b [/tex]


we know that

If [tex] (x-2) [/tex] is a factor of [tex] x^2 - bx + b [/tex]

then

for [tex] x=2 [/tex]

[tex] x^2 - bx + b [/tex] is equal to zero

so

[tex] 2^2 - b*2 + b=0 [/tex]

[tex] 4 - 2b + b=0 [/tex]

[tex] b=4 [/tex]

therefore


the answer is

the value of b is equal to [tex] 4 [/tex]

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.

Find out more at: https://brainly.com/question/2449381

Otras preguntas