Question

Can anyone check my work?

The numbers 1 and -1 are two roots of p(x)=cx³ +ax² + bx + 2c. What's the third root?
a) 0
b) 2
c) -3
d) -2

So, first I did this

c + a + b + 2c = 0
-c + a - b + 2c = 0

a + b + 3c = 0
a - b + c = 0
--------
2a + 4c = 0

Then by the theory of equations I did this
a = -z
b = -1
c = z

I replaced the values in 2a + 4c = 0 and I got z = 0, option a.

Am I correct?

Answer 1

We have
\[cx^3+ax^2+bx+c=0\]
Let's divide whole equation by c, we get
\[x^3+\frac{a}{c}x^2+\frac{b}{c}+2=0\]
@MelindaR one important thing always make the coefficient of highest power term 1, then only you can use theory of equations, now we know that product of three roots is equal to -2, let the third roots be z
so
\[(-1)\times (1)\times z=-2\]
we get
\[z=2\]

Answer 2

Oh, that's what I thought. I knew I had to divide it... Thanks, Ash! Btw, this equation theory is very useful.

Answer 3

welcome @MelindaR , yeah it's great:D always remember to make the coefficient of highest term 1

Answer 4

Ok, I will. Do you remember when you did you learn that theory?

Answer 5

Yeah, when I were 18

Answer 6

K. :P Thx