Help on stupidly difficult polynomial question?explain as well?

Question

A polynomial is given p(x)=x^3-3x+2


x+2 is a factor of p(x)

1) express p(x) as a product of linear factors
a)sketch the graph of p(x). Indicate the co-ordinates of points where the graph intersects the co-ordinate axes. how do you draw a graph of an eqtn?


b) hence, or otherwise, solve the inequality

x^3-3x+2 is less than zero

its got a cubed in how do you solve it?

Answer 1

i dont know

Answer 2

Notice that 1 is a zero of p(x). So x-1 is a factor
By long division or synthetic division,
the quotient is x²+x-2 = (x+2)(x-1).
So the linear factors are (x+2)(x-1)².
To sketch the graph plot many points.
Notice that since x=1 is a double root of p(x)
the graph is tangent to the x-axis at (1,0).
For part b, p(x) = (x+2)(x-1)² <0.
But (x-1)² is always nonnegative,
so the only possibility is x < -2.
You can see this from the graph.
It crosses the x-axis at x = -2 and never
goes below it again.

Answer 3

1) if you divide x^3-3x+2 by x+2, then we get:

x^2-2x+1 which is the same as: (x-1)^2, hence

p(x)=x^3-3x+2= (x-1)^2(x+2)

If we make this equal to zero 0=(x-1)^2(x+2) we can find the x-intersects x=1 x=-2 by making each of the factors equal to zero and solving the equation

You can plot a couple of points in order to get the graph:

we can start by making x=0
p(0)=(0)^3-3(0)+2= 2

making x=2

p(2)=(2)^3-3(2)+2=4

making x=3

p(3)=(3)^3-3(3)+2=20

making x=-1

p(-1)=(-1)^3-3(-1)+2=4

making x=-2

p(-2)=(-2)^3-3(-2)+2=0

making x=-3

p(-3)=(-3)^3-3(-3)+2=-11

At this point you have the following values:

x | -3 | -2 | -1| 0 | 1 | 2 | 3 |
y |-11| 0 | 4 | 2 | 0 | 2 |20|

If you plot this points, you will have an idea as to what the graph looks like

b) p(x)=(x+2) (x-1)^2

(x-1)^2(x+2) < 0

Since (x-1)^2 is a square, it will never be negative and it will be up to the value of (x+2) to make the equation true

x+2 < 0

x< -2

Answer 4

x^3 – 3x + 2
= x^3 – x^2 + x^2 – x – 2x +2
= x^2(x – 1) + x((x – 1) – 2((x – 1)
= (x^2 +x – 2) (x – 1)
= (x^2 – x + 2x – 2)(x – 1)
= [x (x – 1) + 2(x +1)](x – 1)
= (x + 2)(x – 1)(x – 1)
i.e. (x + 2) is a factor of x^3 – 3x + 2

Answer 5

you have to divide p(x)=x^3-3x+2 on x+2 which is going to give you a polynomial of degree 2 then you have to analyze it to its main factors .

to draw the graph you have to find the zeros of the equation then substitute with some random points to help you draw the graph.

Answer 6

a) Use sythetic division to get quotient and then factor,
p(x) = (x-1)^2 (x+2)
To sketch the graph of p(x), you can use x-intercepts, y-intercept, and end behaviors.

b) (x-1)^2 (x+2) < 0 => x < -2, since (x-1)^2 is always positive.

Answer 7

you know dat x+2 is a factor of p(x)
so divide p(x) wid x+2
you will get a quadratic equation
then u can solve da problem