Algebraic Expressions

Identifying components of algebraic expressions

Let’s revisit some of the components of algebraic expressions that we’ve seen in earlier chapters by example.

Component	Algebraic expresison	Example
Variables	$x^2-x-6$x2−x−6	$x$x
Terms	$x^2-x-6$x2−x−6	$x^2$x2, $-x$−x, and $-6$−6
Constant Terms	$x^2-x-6$x2−x−6	$-6$−6
Like Terms	$x^2-x-6+3x^2$x2−x−6+3x2	$x^2$x2 and $3x^2$3x2
Factors	$x^2-x-6$x2−x−6	$x-3$x−3,$x+2$x+2, $x^2-x-6$x2−x−6 and $1$1
Coefficients	$3x^2-x-6$3x2−x−6	$3$3 is a coefficient of $x^2$x2, $-1$−1 is a coefficient of $x$x

 

Question 1

QUESTION 2

question 3

Substituting into algebraic expressions

We often want to substitute values for the variables in an algebraic expression. That way we can evaluate the expression to yield a numerical result. Let’s go through an example below.

Consider the algebraic expression $a^2+2ab-4c^2+\sqrt{a}$a2+2ab−4c2+√a. Let's evaluate this expression when $a=4$a=4, $b=3$b=3 and $c=2$c=2.

Substituting $a=4$a=4, $b=3$b=3 and $c=2$c=2, the expression becomes:

$a^2+2ab-4c^2+\sqrt{a}$a2+2ab−4c2+√a	$=$=	$\left(4\right)^2+2\times\left(4\right)\times\left(3\right)-4\left(2\right)^2+\sqrt{\left(4\right)}$(4)2+2×(4)×(3)−4(2)2+√(4)
	$=$=	$16+24-16+2$16+24−16+2
	$=$=	$26$26

 

question 1

QUESTION 2

question 3

Substituting into formulae or equations

When we have two expressions separated by an equal sign, we have an equation. In the case that our equation contains many variables, we can find the value of one variable by substituting values for the other variables.

Consider the equation of a parabola $y=ax^2+bx+c$y=ax2+bx+c. Find the value of $y$y when $a=1$a=1, $b=4$b=4, $c=4$c=4 and $x=2$x=2.

$y=ax^2+bx+c$y=ax2+bx+c

$y=\left(1\right)\left(2\right)^2+\left(4\right)\times\left(2\right)+\left(4\right)$y=(1)(2)2+(4)×(2)+(4)

$y=4+8+4$y=4+8+4

$y=16$y=16

 

Sometimes the variable of interest isn’t the subject of the equation. In this case, we will have to do some work to rearrange the equation so that the variable becomes the subject.

Consider the equation of a parabola again, $y=ax^2+bx+c$y=ax2+bx+c. Find the value of $x$x when $y=0$y=0, $a=1$a=1, $b=4$b=4 and $c=4$c=4.

$y=ax^2+bx+c$y=ax2+bx+c

$0=\left(1\right)x^2+\left(4\right)x+\left(4\right)$0=(1)x2+(4)x+(4)

$0=x^2+4x+4$0=x2+4x+4

$0=\left(x+2\right)^2$0=(x+2)2

$0=x+2$0=x+2

$x=-2$x=−2

 

Here we've used the fact that we can write $x^2+4x+4$x2+4x+4 as a perfect square, that is $\left(x+2\right)^2$(x+2)2.

 

Consider the equation $y=\left(bx+c\right)\left(dx+e\right)$y=(bx+c)(dx+e). Find the value of $y$y when $bx+c=-4$bx+c=−4 and $dx+e=5$dx+e=5.

$y=\left(bx+c\right)\left(dx+e\right)$y=(bx+c)(dx+e)

$y=-4\times5$y=−4×5

$y=-20$y=−20

$x=-2$x=−2

 

Here we've used the fact that we can replace a whole algebraic expression with a given numerical value.

 

Some further examples are given below.

 

question 1

question 2

question 3