Use the Factor Theorem to show that x-1 is a factor of?

Question

P(x)=x^3 - 3 x^2 + 10 x - 8

The function value P(1)= .
Thus, x-1 is a ________ of P(x)

Answer 1

P(1) = 1^3 - 3(1^2) +10 (1) -8
= 1 - 3+10 - 8
= 0
no remainder when P(x) is divided by x-1
therefore x-1 is a factor of P(x)

Answer 2

Using The Factor Theorem

Answer 3

Saying (x-1) is a factor basically means substitute a 1 everytime you see an x in the equation!. To prove it is a factor, the answer has to equal 0.
If it was (x + 2), -2 written instead of x would have to give you 0. But for your example, (x-1) 1 instead of x should give you 0. I'll show you how.
(x-1) into P(x) = x^3 - 3 x^2 + 10 x - 8
gives P(1) = 1^3 - 3 x 1^2 + 10x1 - 8
P(1) = 1 - 3 + 10 - 8 = 0
Ta da! This proves that when you insert a 1 everytime you see an x, you got an answer of 0 meaning it is a factor. When x = 1,
x-1 = 0
so (x-1) is a factor! hope iv helped x

Answer 4

What you can do is to sub "1" for all of your x values. If your answer is zero then (X-1) is a factor of that equation. If is doesn't then it is not.

Ex: (1)^3-3(1)^2+10(1)-8=???

or you can use a graphing calc and enter the equation and go to table to see when x=1 what does y=??

Answer 5

if x-1 is a factor of x^3-3x^2+10x-8 then when x=1 is plugged into the equation the result should be zero. so:

1^3-3(1)^2+10(1)-8=0

Answer 6

P(1) = 0.
Thus x-1 is a factor of P(x).

Answer 7

by using fact of x=a million is fulfill in this equation so (x-a million) is afactor of the poynomial. additionally, x^3+3x^2-2x-2=(x-a million)(x^2+4x+2)=0 then x=a million, x=-2+(2)^0.5, x=-2-(2)^0.5 are the roots of equation.