using the factor theorem to show that x-c is a factor of P(X) for the given values of c, and factor
1. P(x)=x3-5x^2+8x-4; c=1
x^3 -5x^2 +8x -4 = (x-1)(x^2 -4x +4)
= (x-1)(x-2)^2
To use the factor theorem to show that x-c is a factor of P(x) for the given value of c, we need to check if P(c) equals zero.
Let's substitute c=1 into P(x):
P(1) = (1)^3 - 5(1)^2 + 8(1) - 4
= 1 - 5 + 8 - 4
= 0
Since P(1) equals zero, we can conclude that x-1 is a factor of P(x).
To use the factor theorem to show that x-c is a factor of P(x), you need to check if P(c) equals to 0. If P(c) equals 0, it means that (x-c) is a factor of P(x).
In this case, we have P(x) = x^3 - 5x^2 + 8x - 4, and c = 1.
Step 1: Calculate P(c)
Replace every 'x' in P(x) with the value of c:
P(1) = (1)^3 - 5(1)^2 + 8(1) - 4
P(1) = 1 - 5 + 8 - 4
P(1) = 0
Step 2: Check if P(c) = 0.
Since P(1) equals 0, it means that (x-1) is a factor of P(x) when c=1.
Therefore, x-1 is a factor of P(x), or P(x) can be factored as (x-1)(x^2 - 4x + 4).