MCR3U Equivalent Algebraic Expressions: Notes and Practice

Two expressions can look completely different on the page and still represent exactly the same relationship. $2 (x - 3)^{2} + 5$ and $2 x^{2} - 12 x + 23$ look nothing alike, yet they produce identical outputs for every single input. Being able to prove — not just guess — that two expressions are equivalent ties together everything you know about expanding and factoring. It is also the reason the rest of this unit works: switching a quadratic between standard, factored and vertex form, simplifying a rational expression, or tidying a radical are all ways of replacing one expression with an equivalent one that is easier to use. If a step ever produced a non-equivalent expression, every answer after it would be wrong, so knowing how to check equivalence is how you catch your own algebra errors.

Fig 1Standard, factored, and vertex form are all equivalent expressions for the same parabola — they expand to identical polynomials.

What 'equivalent' means

Two algebraic expressions are equivalent if they produce the same output for every value of the variable(s) — not just for one convenient test value, but for all of them. Graphically, their graphs are the same curve, lying exactly on top of each other. Compare this with an equation such as $x^{2} - 16 = x + 4$ , which is true only for particular values ( $x = 5$ or $x = - 4$ ), where two different graphs happen to cross. The reliable way to prove equivalence is to fully expand and simplify both expressions and check that the results match term for term: the same coefficient on $x^{2}$ , the same on $x$ , and the same constant. To prove two expressions are not equivalent, a single value where they disagree (a counterexample) is enough.

One matching value is not proof

Two genuinely different expressions can still agree at a particular

x

-value by coincidence (their graphs might just cross there). Testing a single value can rule expressions out (if they disagree, they're definitely not equivalent) but can never confirm equivalence on its own — full expansion is the only proof. For example,

x^{2}

and

3 x - 2

both equal

1

x = 1

(and

4

x = 2

), but at

x = 0

they give

0

and

- 2

Expanding with the area model

Expanding means removing every bracket using the distributive property: every term in the first bracket multiplies every term in the second. An area model makes this impossible to miss. For $(2 x + 3) (x - 4)$ , draw a rectangle with side lengths $2 x + 3$ and $x - 4$ and split it into a grid, one row per term of the first factor and one column per term of the second. Each cell is one product: $2 x \cdot x = 2 x^{2}$ , $2 x \cdot (- 4) = - 8 x$ , $3 \cdot x = 3 x$ and $3 \cdot (- 4) = - 12$ . The total area is the sum of the cells, $2 x^{2} - 8 x + 3 x - 12 = 2 x^{2} - 5 x - 12$ . The two $x$ -cells, $- 8 x$ and $3 x$ , are like terms, and they combine into the middle term.

Fig 2An area model for (2x+3)(x−4): each cell is one term times one term, and like terms are collected at the end.

The same grid handles longer products. A binomial times a trinomial is a $2$ -by- $3$ grid, so there are $6$ products before collecting: $(x + 2) (x^{2} - 3 x + 1) = x^{3} - 3 x^{2} + x + 2 x^{2} - 6 x + 2 = x^{3} - x^{2} - 5 x + 2 .$ For three binomials, expand two of them first, then multiply the result by the third. Counting cells is a useful check: if a $2$ -by- $3$ product gives you only $5$ terms before collecting, one product is missing. Often you only need one coefficient. The $x^{2}$ term in $(x + 2) (x^{2} - 3 x + 1)$ comes only from $x \cdot (- 3 x)$ and $2 \cdot x^{2}$ , so its coefficient is $- 3 + 2 = - 1$ , with no need to expand the rest.

VideoKhan Academy multiplies binomials with an area model, the same grid this section uses so that every term in one bracket meets every term in the other.Watch on YouTube

Special products and simplifying

Pattern	Expansion	Example
$(A + B)^{2}$	$A^{2} + 2 A B + B^{2}$	$(x + 5)^{2} = x^{2} + 10 x + 25$
$(A - B)^{2}$	$A^{2} - 2 A B + B^{2}$	$(2 x - 1)^{2} = 4 x^{2} - 4 x + 1$
$(A + B) (A - B)$	$A^{2} - B^{2}$	$(x + 6) (x - 6) = x^{2} - 36$

A squared binomial always has a middle term; only the product of conjugates loses it.

Why does $(A + B)^{2}$ have a middle term? Because it means $(A + B) (A + B)$ , a $2$ -by- $2$ grid with two cells equal to $A B$ . That is why $(x - 4)^{2} = x^{2} - 8 x + 16$ , not $x^{2} - 16$ . Three habits prevent most errors. First, square before you multiply by a coefficient: $2 (x - 3)^{2}$ means $2 (x^{2} - 6 x + 9) = 2 x^{2} - 12 x + 18$ , not $(2 x - 6)^{2}$ . Second, when subtracting a bracket, the negative reaches every term: $- (x^{2} - 4 x - 5) = - x^{2} + 4 x + 5$ . Third, like terms must have the same variable and the same exponent, so $3 x^{2}$ and $3 x$ never combine. Simplifying means collecting all like terms, and both expressions must be fully simplified before you compare them.

Proving or disproving equivalence

Quick test: substitute an easy value such as x = 0 or x = 1 into both. If the outputs differ, stop: they are not equivalent.
Otherwise expand and simplify the first expression completely, one bracket at a time.
Expand and simplify the second expression completely.
Write both in descending powers of $x$ and compare coefficients term by term.
State the conclusion in words: "equivalent, since both simplify to $2 x^{2} - 5 x - 12$ " or "not equivalent, since at $x = 0$ they give $16$ and $- 16$ ".

Graph as a check

With graphing technology, enter both expressions as

y_{1}

and

y_{2}

. Equivalent expressions give one curve drawn twice; non-equivalent ones separate somewhere. A graph is strong evidence, but the written expansion is still the proof. The same idea works in a table of values: equivalent expressions give identical output columns for every input you try, while non-equivalent ones may agree at a value or two and then split apart, which is exactly why one matching value never settles the question.

Equivalent Algebraic Expressions

Key things to remember

What 'equivalent' means

Expanding with the area model

Special products and simplifying

Proving or disproving equivalence

Worked examples

Where marks get lost

Flashcards 9

Practice