Question

Please check:
-3x^3-7x^2-3x-7 need to express as a product of linear factors and find all zeros

X^2(-3x-7)(-3x-7)
Zeros are -7/3

Answer 1

this is an algebra 1 function, no other technics need to be applied.

-3x^3-7x^2-3x-7=0
-3x^3-3x-7x^2-7=0
-3x(x^2-1)+(-7x)(x^2-1)=0
(-3x-7x)(x^2-1)=0
(-10x)(x^2-1)=0

-10x=0
x=0

x^2-1=0
x^2=1
x= -1, 1

Answer 2

So the zeros are

0, 1, -1

the y intercept is just going to be the last constant, in this case -7.

So zeros, 0, -1, 1
y-intercept, -7

Answer 3

I don't quite follow how you took out the 3x and got -3x(x^2-1)+(-7x)(x^2-1)

Answer 4

I factored it out, that's all.

Answer 5

I mean not factored OUT, I just factored.

Answer 6

Shouldn't I factor -3x^3-7x^2 as x^2(-3x-7) then have (-3x -7) as the other factor?

Answer 7

you would get the same thing.

Answer 8

How do I then get from (-3x-7) to the zeros of 0,-1,1?

Answer 9

\[ -3x^3-7x^2-3x-7 \\ x^2(-3x -7) + (-3x-7) \\ (x^2 +1)(-3x-7) =0\]

Answer 10

the term
-3x -7=0 gives us
-3x = 7
x = -7/3

Answer 11

but the
x^2 + 1 =0
gives us
x^2 = -1

the square root of -1 is an imaginary number ... you get ± i
(where "i" is short for sqr(-1) )

Answer 12

If you are not studying complex roots, this might mean there is a typo in the original equation

Answer 13

So I was on the right track with the -7/3 but didn't go far enough with the x^2+1?

Answer 14

you should have gotten to this
\[ x^2(-3x -7) + (-3x-7) \]
to make it a bit more obvious, this is like factoring
\[ x^2 A + A = (x^2 +1) A \]

Answer 15

because minus signs are notorious for causing problems, we could at the very start
\[ -3x^3-7x^2-3x-7= 0 \]
multiply both sides by -1 (all terms!) to get
\[ 3x^3+7x^2+3x+7=0 \]
we will still get the same answer.... but it's less confusing (for me at any rate)

Answer 16

I started out doing that then was confused by the reply. So. My zeros are -7/3, -1,1