How do you factor y= x^3-2x^2+x-2y=x3−2x2+x−2?

3 Answers
Jim G.
Jun 13, 2018

(x-2)(x+i)(x-i)(x−2)(x+i)(x−i)

Explanation:

"factor the terms "color(blue)"by grouping"factor the terms by grouping

=color(red)(x^2)(x-2)color(red)(+1)(x-2)=x2(x−2)+1(x−2)

"take out the "color(blue)"common factor "(x-2)take out the common factor (x−2)

=(x-2)(color(red)(x^2+1))=(x−2)(x2+1)

x^2+1" can be factored using "color(blue)"difference of squares"x2+1 can be factored using difference of squares

a^2-b^2=(a-b)(a+b)a2−b2=(a−b)(a+b)

"with "a=x" and "b=ito(i=sqrt-1)with a=x and b=i→(i=√−1)

=(x-2)(x+i)(x-i)larrcolor(red)"in factored form"=(x−2)(x+i)(x−i)←in factored form

anskat
Jun 13, 2018

y=(x-2) xx (x^2+ 1)y=(x−2)×(x2+1)

Explanation:

x^3 -2x^2x3−2x2 can be written as
x^2xx(x-2)x2×(x−2), so we can re-write the original equation as

y=x^2xx(x-2) + (x-2)y=x2×(x−2)+(x−2)
or
y=x^2xx(x-2) + 1xx(x-2)y=x2×(x−2)+1×(x−2)

and bringing the common factor (x-2)(x−2) to the front, gives the factors:

y=(x-2) xx (x^2+ 1)y=(x−2)×(x2+1)

solving the equation for y=0y=0, gives the solution x=2x=2, meaning that the graph of the equation crosses the x-axis at the point (2,0)

solving the equation for x=0x=0 gives y=-2y=−2, meaning that the graph of the equation crosses the y-axis at (0,-2)

graph{x^3-2x^2+x-2 [-10, 10, -5, 5]}

Boba
Jun 13, 2018

y=(x^2+1)(x-2)y=(x2+1)(x−2)

Explanation:

y=x^3-2x^2+x-2y=x3−2x2+x−2

We will use a factoring method called grouping. If we group the first and the third and the second and the fourth together, each grout has its own greatest common factor:

y=x(x^2+1)-2(x^2+1)y=x(x2+1)−2(x2+1)

Now we see another CF between the two terms :x^2+1x2+1. We can factor it out.

y=(x^2+1)(x-2)y=(x2+1)(x−2)

Related questions
See all questions in Factor Polynomials Using Special Products
Impact of this question
7298 views around the world
You can reuse this answer
Creative Commons License