Question

How to find all real or imaginary zeros for each polynomial function:

F (x) = 2x^3-4x^2+3x

F (x)= x^3+2x^2-6x

I know how to do these type of problems but, both do not have a solution found in the given P/Q set.... so, I do not know what to do... help! (This was not covered in lecture & exam is tomorrow !)

Answer 1

\(\large\color{black}{ \displaystyle f(x)=2x^3-4x^2+3x }\)

Answer 2

To find the zero, set:

\(\large\color{black}{ \displaystyle 0=2x^3-4x^2+3x }\)

Answer 3

\(\large\color{black}{ \displaystyle 0=x(2x^2-4x+3) }\)

Answer 4

So you have two cases:

[Case 1]
\(\large\color{black}{ \displaystyle x=0 }\)

[Case 2]
\(\large\color{black}{ \displaystyle x=2x^2-4x+3 }\)

Answer 5

Ok am following so far....

Answer 6

Ok, the first case is already a solution,
and case 2 can be solved using the quadratic formula.

Answer 7

So we, factor further right ? (The second case)

Answer 8

corrrect

Answer 9

Ok let me try......

Answer 10

You can't really factor the second case, but you can use the quadratic formula.

Answer 11

\(\bbox[8pt, yellow ,border:8px solid black]{\Huge \Huge{x=~} \Huge{ \frac{-\color{magenta}{b} \pm\sqrt{ \color{magenta}{b} ^2-4 \color{blue}{a} \color{red}{c}}}{2 \color{blue}{a}} } ~ }\)

when the equation is \(\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\).

Answer 12

Oh, I remember now.... ok let me try that! Once I sove I have my answer ?

Answer 13

Ahh.... I think am doing it wrong... well I have

Answer 14

4+/- (square root of) -8 (over )4

Answer 15

@SolomonZelman

Answer 16

@triciaal

Answer 17

\[F (x)= x^3+2x^2-6x=x(x^2+2x-6)\] one zero is 0, the other you get from solving \[x^2+2x-6=0\]

Answer 18

or are you on the first one

Answer 19

Either is fine with me, let me try to do what you said....so to solve for x ?

Answer 20

yes
2 is even so skip the quadratic formula

Answer 21

easier to complete the square do you don't have to rewrite in simplest radical form and cancel

Answer 22

it it is totally unclear what i am saying, let me know
that is for solving \[x^2+2x-6=0\]

Answer 23

Would it be +/- (square root of ) 3 ? Or was I not suppose to drop off that other x ?

Answer 24

i don't think so no

Answer 25

\[x^2+2x-6=0\\
x^2+2x=6\] is a start

Answer 26

now what is half of 2?

Answer 27

1

Answer 28

and 1 squared?

Answer 29

Still 1 ?.....

Answer 30

Unless if negative the i

Answer 31

yes so go right from \[x^2+2x=6\] so \[(x+1)^2=6+1\\
(x+1)^2=7\]

Answer 32

that means \[x+1=\pm\sqrt7\\
x=-1\pm\sqrt7\]

Answer 33

no quadratic formula was hurt in this process
you would get the same answer, but it would take a lot more work to get it

Answer 34

Oh I see, yes.... my professor definitely didn't go over this! And yes this seems simpler

Answer 35

Thank you ! 😁 @satellite73