Question

A little help here please? :( The function of the polynomial is x^3-2x^2-4x+8 and its zero is 2, -2, and 2. And the degree of P(x) is 3. I don't understand how they got the zeros (2, -2 and 2) please explain me this. This is for my report this coming friday. :(

Answer 1

\[(x-2)(x-2)(x+2)\]

Answer 2

given th zeros we can find the factored form of the function as satellite as done.

solving (x - 2)(x - 2)(x + 2) = 0 gives the zeros.

the above would have been found by factoring
x^3-2x^2-4x+8

Answer 3

using the rationalr oot theorem its a good better that 1 or more zeros are 2 because factors of the last term 8 are 2 and 4

find f(2) to confirm this (by checking if f(2) = 0)

Answer 4

so one of the facors is (x-2)
now divide f(x) by x-2 using long or synthetic division
finally factorize the quotient

Answer 5

* a good bet

Answer 6

I think your are explaining me the other when the zeros are guven the remaining zeros can be found. I'm asking, how 2, -2, and 2 was got by the polynomial function P(x)=x^3-2x^2-4x+8 @cwrw238 :)


@satellite73 : is that the factor of x^3-2x^2-4x+8 ? :)

Answer 7

\[\Large x^3-2x^2-4x+8=x^2(x-2)-4(x-2)\\
\Large (x-2)(x^2-4)=0\\
\Large => x-2=0, or,x^2-4=0\\
\Large => x=2,or x=\mp2\]

Answer 8

Hmm. Thank yooou so much @cinar , now i understand :)

Answer 9

by the way, why is it ∓2? @cinar

Answer 10

@campbell_st : Please, let me understand why is it ∓2? in the solution if @cinar

Answer 11

it is +/- because you have x^2 -4 = 0 by null factor. and therefore x = sqrt 4
sqrt 4 is posotibve or negative 2
(-2)^2 = 4 just as 2^2 = 4

Answer 12

ok... so here is another slant on the above answers...

the zeros or roots of a polynomial are where the polynomial cuts the x-axis

so if your polynomial is

\[P(x) = x^3 -2x^2 - 4x + 8\]

you are told the zeros... to show it substitute x = 2 into your polynomial

\[P(2) = (2)^2 -2(2)^2 -4(2) + 8\]

calculate it out and you get zero...

so when x = 2 P(2) = 0

hence the name...

hope this makes sense

you can substitute x = -2 and find P(-2) = 0 ... as well

Answer 13

and as for the solutions... there are only 2, x = 2 and x = -2... the 2nd solution is repeated.... (x-2)(x-2) = (x-2)^2

this is normally described as multiplicity of 2.... the zero at x = -2 occurs twice

Answer 14

@campbell_st I don't think she was given the zeros, I think she has a problem and answer and wants to understand how to get there.

Like @cinar showed, this polynomial can conveniently be factored by grouping. That might be as far as you've gotten in this lesson.

But in general, you can use the rational root theorem as was discussed above, OR a graphing calculator if allowed, to find a first rational root. Then pull out the factor associated with that root by synthetic division, and what you have left is a quadratic. Solve using methods for solving quadratic equations and that will give you the other 2 roots.

Answer 15

@DebbieG : debbieg's right. I need to understand how to get there of the answer. :3 cause i don't get it. Why in the solution of @cinar is ∓2? why not positive 2? :3

Answer 16

Because once you reach this point:

\(\Large (x-2)(x^2-4)=0\\ \)

Now you have to factor the \(\Large (x^2-4)\\ \)

That's a difference of squares. What do you get when you factor it?

Answer 17

You get 2 roots there - so he is just combining them using \(\pm\) notation. You could write them each separately too. {2, -2}

Of course, the 2 is already a root of the polynomial, so that one is a repeated root.

Answer 18

When all is said and done, what you end up with is:

\(\Large y= (x-2)^2(x+2)\\ \)

Answer 19

Ahhhh. :) nooow, i understand. Thank yooooou so much @DebbieG your really helped me a lot. ^_^v and thanks also to @campbell_st for sharing your idea and @cinar for sharing the solution. Thank yoooooou so much! :)

Answer 20

You're welcome. :)