# Simplify (2x^2+3x+4)(5x^2+7x+6)?

May 8, 2016

$\left(2 {x}^{2} + 3 x + 4\right) \left(5 {x}^{2} + 7 x + 6\right) = 10 {x}^{4} + 29 {x}^{3} + 53 {x}^{2} + 46 x + 24$

#### Explanation:

Since the coefficients are all positive, then if you have a calculator to hand we can calculate the coefficients of the product like this:

$20304 \times 50706 = 1029534624$

How does that help?

Each pair of digits represents a coefficient, so:

$\textcolor{p u r p \le}{2} \textcolor{b l u e}{03} \textcolor{g r e e n}{04} \times \textcolor{p u r p \le}{5} \textcolor{b l u e}{07} \textcolor{g r e e n}{06} = \textcolor{red}{10} \textcolor{m a \ge n t a}{29} \textcolor{p u r p \le}{53} \textcolor{b l u e}{46} \textcolor{g r e e n}{24}$

represents:

$\left(\textcolor{p u r p \le}{2} {x}^{2} + \textcolor{b l u e}{3} x + \textcolor{g r e e n}{4}\right) \left(\textcolor{p u r p \le}{5} {x}^{2} + \textcolor{b l u e}{7} x + \textcolor{g r e e n}{6}\right) = \textcolor{red}{10} {x}^{4} + \textcolor{m a \ge n t a}{29} {x}^{3} + \textcolor{p u r p \le}{53} {x}^{2} + \textcolor{b l u e}{46} x + \textcolor{g r e e n}{24}$

This is essentially putting $x = 100$ in each of the expressions.

Since the largest result we can get by multiplying a one digit number by a one digit number is less than $100$ - that is $9 \times 9 = 81 < 100$ - there will be no carries to mess up the pairs of digits in the answer.

May 8, 2016

$\left(2 {x}^{2} + 3 x + 4\right) \left(5 {x}^{2} + 7 x + 6\right) = 10 {x}^{4} + 29 {x}^{3} + 53 {x}^{2} + 46 x + 24$

#### Explanation:

The formal way of doing the multiplication is something like this:

$\left(2 {x}^{2} + 3 x + 4\right) \left(5 {x}^{2} + 7 x + 6\right)$

$= 2 {x}^{2} \left(5 {x}^{2} + 7 x + 6\right) + 3 x \left(5 {x}^{2} + 7 x + 6\right) + 4 \left(5 {x}^{2} + 7 x + 6\right)$

$= \left(\left(2 \cdot 5\right) {x}^{4} + \left(2 \cdot 7\right) {x}^{3} + \left(2 \cdot 6\right) {x}^{2}\right) + \left(\left(3 \cdot 5\right) {x}^{3} + \left(3 \cdot 7\right) {x}^{2} + \left(3 \cdot 6\right) x\right) + \left(\left(4 \cdot 5\right) {x}^{2} + \left(4 \cdot 7\right) x + \left(4 \cdot 6\right)\right)$

$= \left(10 {x}^{4} + 14 {x}^{3} + 12 {x}^{2}\right) + \left(15 {x}^{3} + 21 {x}^{2} + 18 x\right) + \left(20 {x}^{2} + 28 x + 24\right)$

$= 10 {x}^{4} + \left(14 + 15\right) {x}^{3} + \left(12 + 21 + 20\right) {x}^{2} + \left(18 + 28\right) x + 24$

$= 10 {x}^{4} + 29 {x}^{3} + 53 {x}^{2} + 46 x + 24$

A more compact way of finding the coefficients uses a grid to help us find the combinations of terms to add:

$\underline{\textcolor{w h i t e}{000} | \textcolor{w h i t e}{00} 5 \textcolor{w h i t e}{00} 7 \textcolor{w h i t e}{00} 6 \textcolor{w h i t e}{00}}$
$\textcolor{w h i t e}{000} | \textcolor{w h i t e}{0}$
$\textcolor{w h i t e}{0} 2 \textcolor{w h i t e}{0} | \textcolor{w h i t e}{0} \textcolor{red}{10} \textcolor{w h i t e}{0} \textcolor{m a \ge n t a}{14} \textcolor{w h i t e}{0} \textcolor{p u r p \le}{12}$
$\textcolor{w h i t e}{000} | \textcolor{w h i t e}{0}$
$\textcolor{w h i t e}{0} 3 \textcolor{w h i t e}{0} | \textcolor{w h i t e}{0} \textcolor{m a \ge n t a}{15} \textcolor{w h i t e}{0} \textcolor{p u r p \le}{21} \textcolor{w h i t e}{0} \textcolor{b l u e}{18}$
$\textcolor{w h i t e}{000} | \textcolor{w h i t e}{0}$
$\textcolor{w h i t e}{0} 4 \textcolor{w h i t e}{0} | \textcolor{w h i t e}{0} \textcolor{p u r p \le}{20} \textcolor{w h i t e}{0} \textcolor{b l u e}{28} \textcolor{w h i t e}{0} \textcolor{g r e e n}{24}$

Each of the numbers in the body of the grid is the product of its row label and column label, e.g. $2 \times 7 = \textcolor{m a \ge n t a}{14}$

Then add the reverse diagonals to find:

$\left(2 {x}^{2} + 3 x + 4\right) \left(5 {x}^{2} + 7 x + 6\right)$

$= \textcolor{red}{10} {x}^{4} + \textcolor{m a \ge n t a}{\left(14 + 15\right)} {x}^{3} + \textcolor{p u r p \le}{\left(12 + 21 + 20\right)} {x}^{2} + \textcolor{b l u e}{\left(18 + 28\right)} x + \textcolor{g r e e n}{24}$

$= \textcolor{red}{10} {x}^{4} + \textcolor{m a \ge n t a}{29} {x}^{3} + \textcolor{p u r p \le}{53} {x}^{2} + \textcolor{b l u e}{46} x + \textcolor{g r e e n}{24}$