2.03 Multiplying polynomials

Multiplying two binomials

Recall that to distribute an expression like $2\left(x-3\right)$2(x−3)we use the distributive property: $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC

Now we want to look at how to multiply two binomials together, such as $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d). 

A visual interpretation

Let's look at $\left(x+5\right)\left(x+2\right)$(x+5)(x+2) for example and see how this distribution works visually before we look at the algebraic approach. We will use the area of a rectangle. We can consider the expression $\left(x+5\right)\left(x+2\right)$(x+5)(x+2) to represent the area of a rectangle with side lengths of $x+5$x+5 and $x+2$x+2 as seen below.

Another way to express the area would be to split the large rectangle into four smaller rectangles. Since the area of the whole rectangle is area of the sum of its parts we get the following:

$\left(x+5\right)\left(x+2\right)$(x+5)(x+2)	$=$=	$x^2+2x+5x+10$x2+2x+5x+10
	$=$=	$x^2+7x+10$x2+7x+10

We don't want to have to draw a rectangle every time, so below we'll look at the algebraic approach.

 

An algebraic approach

When we multiply binomials of the form $\left(ax+b\right)\left(cx+d\right)$(ax+b)(cx+d) we can treat the second binomial $\left(cx+d\right)$(cx+d) as a constant term and apply the distributive property in the form $\left(B+C\right)\left(A\right)=BA+CA$(B+C)(A)=BA+CA. The picture below shows this in action:

As you can see in the picture, we end up with two expressions of the form $A\left(B+C\right)$A(B+C). We can distribute these using the distributive property again to arrive at the final answer:

$\left(ax+b\right)\left(cx+d\right)$(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

 

 

We can actually be even more efficient in our algebraic approach by taking a look at the step below.

$\left(ax+b\right)\left(cx+d\right)=acx^2+adx+bcx+bd$(ax+b)(cx+d)=acx2+adx+bcx+bd

Notice that we have multiplied every term in the first bracket by every term in the second bracket. In general, that is what is required to multiply two polynomials together. We often use arrows as shown below to help us get every pair.

By distributing in this way we will get the result $x^2+2x+5x+10=x^2+7x+10$x2+2x+5x+10=x2+7x+10, the same result we would have obtained using the previous method. You may prefer to use this alternate method since it is more efficient.

 

Worked example

Question 1

Distribute and simplify $\left(x-3\right)\left(x+4\right)$(x−3)(x+4) .

Think: We need to multiply both terms inside $\left(x-3\right)$(x−3) by both terms inside $\left(x+4\right)$(x+4).

Do:

$\left(x-3\right)\left(x+4\right)$(x−3)(x+4)	$=$=	$x\left(x+4\right)-3\left(x+4\right)$x(x+4)−3(x+4)
	$=$=	$x^2+4x-3x-12$x2+4x−3x−12	We can jump right to this step using the short-cut mentioned above
	$=$=	$x^2+x-12$x2+x−12

 

Practice questions

Question 2

Question 3

Question 4

 

Multiplying two polynomials

What we know about multiplying two binomials, can be extended to multiply and two polynomials. 

 

Worked example

Question 5

Distribute and simplify $\left(x^3-4\right)\left(x^2+3x+5\right)$(x3−4)(x2+3x+5). What do you notice about the degree of the new polynomial?

Think: We need to multiply every pair of terms between the two parentheses. Then we will distribute and simplify.

Do:

$\left(x^3-4\right)\left(x^2+3x+5\right)$(x3−4)(x2+3x+5)	$=$=	$x^3\left(x^2+3x+5\right)-4\left(x^2+3x+5\right)$x3(x2+3x+5)−4(x2+3x+5)
	$=$=	$x^3\times x^2+x^3\times3x+x^3\times5-4x^2-4\times3x-4\times5$x3×x2+x3×3x+x3×5−4x2−4×3x−4×5
	$=$=	$x^5+3x^4+5x^3-4x^2-12x-20$x5+3x4+5x3−4x2−12x−20

Reflect: When we multiplied a quadratic (degree $2$2) and a cubic (degree $3$3) we ended up with a polynomial with degree$5$5. In general, leading terms of the product will be the product of the leading terms. This means using our laws of exponents, the degree will be the sum of the degrees of the two polynomials. 

 

 

Practice questions

Question 6

Question 7