Question

HELP! Explain the steps to me please.
I know the first step is to distribute the equation on the right.

If 6x^2 + x - c = (cx - 1)(3x + 2), what is the value of c?
F. -2
G. -1
H. 2
J. 3
K. 6

Answer 1

\[6x^2 + x - c = (cx - 1)(3x + 2)\]

Answer 2

So go ahead and distribute

Answer 3

\[6x^2 + x - c = 3cx^2 + 2cx - 3x - 2\]

Answer 4

Combine like terms.

Answer 5

\[6x^2 + 4x - c = 2cx - 2\]

Answer 6

You can factor out c, move the -c on the right I guess

Answer 7

So...
\[6x^2 + 4x = 2cx + c - 2\]

What's next?

Answer 8

move the -2 to the left

Answer 9

factor out c

Answer 10

solve for c

Answer 11

\[6x^2 + 4x + 2 = 2cx + c\]

\[6x^2 + 4x + 2 = c(2x + 1)\]

Answer 12

(Divide both sides by (2x+1)

Answer 13

Oh boy... I just learned this in precalc. I hate this topic although I got a 95.

Answer 14

d00d no need to distribute
u know that the 6x^2 term will be made when 3x will get multiplied to cx

so
\[3x \times cx=6x^2\]
x=2 (;

Answer 15

\(6x^2+x-c=(cx-1)(3x+2)\)
\(6x^2+x-c = 3cx^2+2cx-3x-2\)
\(6x^2+x-c - 3cx^2-2cx+3x+2=0\)
\(6x^2+4x+2-3cx^2-2cx-c=0\)
\(6x^2+4x+2=3cx^2+2cx+c\)
\(6x^2+4x+2=c(3x^2+2x+1)\)
\[c = \frac{ 6x^2+4x+2 }{ 3x^2+2x+1 }\] sorry I'm slow with latex

Answer 16

Changing all the square brackets to round

Answer 17

I sort of overcomplicated this I guess, but yeah x = 2 -.- qwerty

Answer 18

Astro, I understand it a little better now.
Qwerty, can you explain your method a little clearer?

Answer 19

All methods work

Answer 20

\(\Huge\ddot\smile\) mmk

Answer 21

\[c = \frac{ 6x^2+4x+2 }{ 3x^2+2x+1 } \implies \frac{ 2(3x^2+2x+1) }{ (3x^2+2x+1) } = 2\]

Answer 22

so we have this- \(6x^2+x-c=(cx-1)(3x+2)\)

We gotta prove that the left hand side is equal to the right hand side

now we got only 1term of the degree 2 on the Left and it is \(\huge\color{red}{6x^2}\) and because both sides gotta be equal we need a 6x^2 on the right as well

now look at this-> right hand side->\(\large\color{violet}{(cx-1)(3x+2)}\)

now when we open the brackets up we need to get a \(\color{red}{6x^2}\) which is a 2nd degree term

a second degree term is obtained when we multiply a 1st degree term with another 1st degree term

now look at this->\((\color{blue}{cs}-1)(\color{blue}{3x}+2)\)

here \(\color{gold}{cs~and~3x}\) are the only 1st degree terms
multiplying them together we must get a \(\color{red}{6x^2}\) so that LHS=RHS

so

\(\large\color{black}{cs \times 3x=6x^2}\)
\(\large\color{black}{c=2}\)

Answer 23

I understand it now. Thank you to all!

Answer 24

ayy np (;