Question

solve plzzz
ƒ(x) = x4 - 7x2 + 6x

Answer 1

\[ƒ(x) = x^4 - 7x^2 + 6x\]

Answer 2

to find the zeros of it, set it equal zero and solve for x.
\[0=x^4-7x^2+6x\]factor out of x\[0=x(x^3-7x^2+6)\]
so one of the zeros of the function is\[0\]
now,\[0=x^3-7x^2+6--->x^3-7x^2=-6-->(x^2-7x)x=-6\]
so another zero of the function is\[-6\]
now you have left\[x^2-7x=-6.-->x^2-7x+6=0--->(x-6)(x-1)=0\]
so the remaining two zeros of the function are \[6 & 1\]
so you have the zeros of the function, they are,
\[0,-6,6,1\]
@dan815 @╰☆╮Openstudier╰☆╮@Callisto
Am I right?

Answer 3

idk what exactly do you mean by solve.

Answer 4

I found the zeros of the function.

Answer 5

i think this is wrong

Answer 6

Yeah, i am not sure..... I tagged but no one is coming, I want to see how to do this too.

Answer 7

solve what?

Answer 8

Dan815 not yet

Answer 9

@skullpatrol, @dan815 Did I find the zeros of the function correctly?

Answer 10

no they are off

Answer 11

can you show...