# How do you find the product of (x + 1) (x^2 + x + 1)?

Jun 13, 2017

See a solution process below:

#### Explanation:

To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(\textcolor{red}{x} + \textcolor{red}{1}\right) \left(\textcolor{b l u e}{{x}^{2}} + \textcolor{b l u e}{x} + \textcolor{b l u e}{1}\right)$ becomes:

$\left(\textcolor{red}{x} \times \textcolor{b l u e}{{x}^{2}}\right) + \left(\textcolor{red}{x} \times \textcolor{b l u e}{x}\right) + \left(\textcolor{red}{x} \times \textcolor{b l u e}{1}\right) + \left(\textcolor{red}{1} \times \textcolor{b l u e}{{x}^{2}}\right) + \left(\textcolor{red}{1} \times \textcolor{b l u e}{x}\right) + \left(\textcolor{red}{1} \times \textcolor{b l u e}{1}\right)$

${x}^{3} + {x}^{2} + x + {x}^{2} + x + 1$

We can now group and combine like terms:

${x}^{3} + 1 {x}^{2} + 1 x + 1 {x}^{2} + 1 x + 1$

${x}^{3} + 1 {x}^{2} + 1 {x}^{2} + 1 x + 1 x + 1$

${x}^{3} + \left(1 + 1\right) {x}^{2} + \left(1 + 1\right) x + 1$

${x}^{3} + 2 {x}^{2} + 2 x + 1$

Jun 13, 2017

$\left(x + 1\right) \left({x}^{2} + x + 1\right) = {x}^{3} + 2 {x}^{2} + 2 x + 1$

#### Explanation:

The way I like to do it is longer to explain than to do...

Look at each possible power of $x$ in descending order and add up the different ways of getting it.

So in our example:

Given:

$\left(x + 1\right) \left({x}^{2} + x + 1\right)$

we can tell that the highest possible power of $x$ in the product is $3$, so work down from there:

$\textcolor{w h i t e}{}$
${x}^{3}$: This can only result from multiplying the $x$ in the binomial by the ${x}^{2}$ in the trinomial, so the coefficient is:

$1 \cdot 1 = \textcolor{b l u e}{1}$

So we can start to write:

$\left(x + 1\right) \left({x}^{2} + x + 1\right) = {x}^{3.} . .$

$\textcolor{w h i t e}{}$
${x}^{2}$: This can result from $x \cdot x$ or $1 \cdot {x}^{2}$, so the coefficient is:

$1 \cdot 1 + 1 \cdot 1 = \textcolor{b l u e}{2}$

So we can add $+ 2 {x}^{2}$ to the result:

$\left(x + 1\right) \left({x}^{2} + x + 1\right) = {x}^{3} + 2 {x}^{2.} . .$

$\textcolor{w h i t e}{}$
${x}^{1} :$ This can result from $x \cdot 1$ or $1 \cdot x$, so the coefficient is:

$1 \cdot 1 + 1 \cdot 1 = \textcolor{b l u e}{2}$

So we can add $+ 2 x$ to the result:

$\left(x + 1\right) \left({x}^{2} + x + 1\right) = {x}^{3} + 2 {x}^{2} + 2 x \ldots$

$\textcolor{w h i t e}{}$
${x}^{0} :$ The constant term can only result from multiplying the constant term of the binomial by that of the trinomial, so:

$1 \cdot 1 = \textcolor{b l u e}{1}$

So our final result is:

$\left(x + 1\right) \left({x}^{2} + x + 1\right) = {x}^{3} + 2 {x}^{2} + 2 x + 1$

In practice (and with practice) the result line is all you need write: Adding up the coefficients can be done in your head.