# How do you write y= (x+3) (x+4) (x+5) in standard form?

Apr 11, 2016

$y = {x}^{3} + 12 {x}^{2} + 47 x + 60$

#### Explanation:

The quickest way to multiply this out is probably to look at each power of $x$ in descending order from $3$ to $0$, pick out the choices of terms from each binomial factor that combine to give the power of $x$ and add them together:

${x}^{3}$ : The only way you can get a term in ${x}^{3}$ is by multiplying all of the leading terms of the binomials together: $x \times x \times x$. So the coefficient of ${x}^{3}$ is $\textcolor{b l u e}{1}$.

${x}^{2}$ : The way that you can get a term in ${x}^{2}$ is by choosing one of the binomials to provide a constant term and picking the $x$ term from the other binomials. Hence the coefficient of ${x}^{2}$ in the product of the three binomials is $3 + 4 + 5 = \textcolor{b l u e}{12}$.

$x$ : The way that you can get a term in $x$ is by choosing one of the binomials to provide the $x$ and the other two to provide constant multipliers. Hence the coefficient of $x$ in the product of the three binomials is $4 \cdot 5 + 5 \cdot 3 + 3 \cdot 4 = 20 + 15 + 12 = \textcolor{b l u e}{47}$.

$1$ : The constant term in the product contains no factor of $x$, so the only way you can get it is by multiplying all of the three constants from the binomials together: $3 \cdot 4 \cdot 5 = \textcolor{b l u e}{60}$

Hence we find:

$y = \left(x + 3\right) \left(x + 4\right) \left(x + 5\right) = {x}^{3} + 12 {x}^{2} + 47 x + 60$

I have described the process in a fairly lengthy way, but with practice, you can probably do these sums mostly in your head, gathering together the various combinations of terms.