Question

check 2 x-1 is a factor of p[x]=x^3-x^2+2x-1

Answer 1

Answer:

Step-by-step explanation:

2x - 1 =0

2x = 1

x = 1/2

Now We have to find P(1/2). If P(1/2) is zero, then 2x - 1 is a factor of P(x)

P(x)= x³ - x² + 2x -1

$P(\frac{1}{2})=[\frac{1}{2}]^{3}-[\frac{1}{2}]^{2}+2*[\frac{1}{2}]-1\\\\=\frac{1}{8}-\frac{1}{4}+1-1\\\\=\frac{1}{8}-\frac{1*2}{4*2}+0\\\\=\frac{1}{8}-\frac{2}{8}\\\\=\frac{1-2}{8}\\\\=\frac{-1}{8}\neq 0$

2x - 1 is not a factor of p(x)