When we previously came across binomial products, we learnt that we can use the distributive law twice in order to expand the product into four terms.
$\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d) | $=$= | $ax\left(cx+d\right)+b\left(cx+d\right)$ax(cx+d)+b(cx+d) |
$=$= | $acx^2+adx+bcx+bd$acx2+adx+bcx+bd | |
$=$= | $acx^2+\left(ad+bc\right)x+bd$acx2+(ad+bc)x+bd |
In this case, the variable $x$x appears in both binomial expressions, so we may simplify the expansion further by combining $adx+bcx$adx+bcx into $\left(ad+bc\right)x$(ad+bc)x. We could instead, however, have the following situation.
$\left(ax+b\right)\left(cy+d\right)$(ax+b)(cy+d) | $=$= | $ax\left(cy+d\right)+b\left(cy+d\right)$ax(cy+d)+b(cy+d) |
$=$= | $acxy+adx+bcy+bd$acxy+adx+bcy+bd |
Here you'll notice that we can't simplify the expression any further, since there are no like terms.
We can also expand binomial product, still using the distributive law, by multiplying both terms in the first set of brackets by both terms in the second set of brackets, as shown in the picture below for the product $\left(x+5\right)\left(x+2\right)$(x+5)(x+2)
By expanding in this way we will get the result $x^2+2x+5x+10$x2+2x+5x+10, the same result we would have obtained using the previous method. You may prefer to use this alternate method since it combines two iterations of the distributive law into one line of working.
Now let's have a look at some worked examples where there are different variables in each binomial, or where further simplification is required after expanding.
Expand and simplify the following:
$\left(4r+7\right)\left(7s+2\right)$(4r+7)(7s+2)
Expand and simplify the following:
$\left(2n+5\right)\left(5n+2\right)-4$(2n+5)(5n+2)−4
Expand and simplify:
$6\left(\frac{x}{2}-2\right)\left(x-2\right)+2x$6(x2−2)(x−2)+2x
Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns
Apply algebraic procedures in solving problems