Factor out (m+1) in the following so that the final answer is $\frac{(2m+1) (m+2) (m+1)} {6}$

Question

Question: $\frac{m (m+1) (2m+1) + 6(m+1)^2}{6}$=$\frac{(2m+3)(m+2)(m+1)}{6}$

I must multiply by 6 on both sides and expand the brackets and collect like terms. I'm I correct?

Edit notes: The original problem was posed as: $$ \frac{(m+1) (2m+1) + 6(m+1)^2}{6}= \frac{(2m+3)+(m+2)(m+1)}{6} $$ which did not match the title. The edit provides the corrected question.

Answer 1

The expression in the title can't be the simplified answer, because as a test, if we substitute $m=0$ in your original (you could use any number but $0$ is convenient) we get

$$\dfrac {(2(0)+1)(0+1)+6(0+1)^2}{6}=\dfrac 76$$

Substituting $m=0$ in the expression in the title yields

$$\dfrac {(2(0)+1)(0+2)(0+1)}{6} = \dfrac 26$$

So the $2$ expressions cannot be equal. What you can do to simplify your original problem is the following:

Both terms in the numerator, $(2m+1)(m+1)$ and $6(m+1)^2$, have a common factor of $(m+1)$, so we can factor that out

$$\dfrac {(m+1)[(2m+1)+6(m+1)]}{6}$$

Now you should distribute the $6$ in the numerator and combine like terms

$$\dfrac {(m+1)(2m+1+6m+6)}{6}$$

$$\dfrac {(m+1)(8m+7)}{6}$$

If you want to keep all integers, this is as far as you can go.

Answer 2

The description as given is correct. Multiplying both sides by $6$ yields $$(2m+1) (m+1) + 6(m+1)^2= (2m+3)+(m+2)(m+1)$$ Now what remains is to determine the value of both sides. Let $L_{m}$ be the left and $R_{m}$ be the right. \begin{align} L_{m} &= (2m+1)(m+1)m + 6(m+1)^2 \\ &= (2m^2 + 3m + 1)m + 6(m^2 + 2m +1) \\ &= 2 m^3 + 9 m^2 + 13 m + 6. \end{align} \begin{align} R_{m} &= (2m+3)(m+2)(m+1) \\ &= 2 m^3 + 9m^2 + 13 m + 6. \end{align} This demonstrates that $L_{m} = R_{m}$, as required.