Algebraic Expressions

Stage 8 of 23 Strand 5 of 8 14 lessons

14 illustrated lessons, each teaching the why before the how.

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Substitution

Swap the letter for a number and work it out.

Substituting means putting a number where the letter stands

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3n3n3 lots of n

The expression 3n just means 3 lots of whatever n is.

1234n = 4, so 3n = 12

Let n be 4 and the whole expression becomes 12.

n = 434+ 23 × 4 + 2 = 14

The 4 goes in the box — and only in the box. The 2 waits outside it.

Now you

n = −23−2+ 3

If n = −2, what is 3n + 3?

n = −22−2+ 7

If n = −2, what is 2n + 7?

Like Terms

Only the same letter part can be added.

Only terms with exactly the same letter part can be added together

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3a2a

3 of something and 2 more of the same thing makes 5 of it: 3a + 2a = 5a.

3a2a32a5a

3a and 2a are like terms, so they join into 5a. A 2b would stay apart.

Now you

9x − 5xthe counts subtract

Simplify 9x − 5x

9x − 4xthe counts subtract

Simplify 9x − 4x

Expanding Brackets

Outside multiplies everything inside.

Everything outside a bracket multiplies everything inside it

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A rectangle 3 by 5 has 3 × 5 = 15 squares inside it.

Cut the width into 3 and 2 and the area splits the same way: 3 × (3 + 2) = 3 × 3 + 3 × 2.

axabxbaa(x + b)

The sides can be letters: a(x + b) is the ax strip plus the ab strip.

Now you

x8x

Expand x(x + 8)

−5(x − 7)the −5 multiplies both parts

Expand −5(x − 7)

Factoring by Common Factor

Pull out what both parts share.

Factoring is putting the bracket back by pulling out what both share

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4 × 3 plus 4 × 4, and both lots share the 4.

4a4bab44(a + b)

Take the shared 4 outside and the rest goes into the bracket.

Now you

10x152x35

Factor 10x + 15

18x243x46

Factor 18x + 24

Factoring by Grouping

Four terms, taken two at a time.

Four terms with no common factor can still split into two pairs that share one

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xy + 3x + 2y + 6take the terms two at a time(xy + 3x) + (2y + 6)

No single factor divides all four, so group them two at a time — each pair shares one.

(xy + 3x) + (2y + 6)x out of the first pair, 2 out of the secondx(y + 3) + 2(y + 3)

Both pairs leave the same bracket behind: y + 3. That is the whole method.

x(y + 3) + 2(y + 3)y + 3 is now a common factor(y + 3)(x + 2)

Two terms, both holding y + 3, so take it out: (y + 3)(x + 2).

ax + bx + kay + kbyfactor each pairx(a + b) + ky(a + b)take out a + b(a + b)(x + ky)

In letters it is the same three moves, and the matching bracket is a + b.

Now you

xy + 2x + 5y + 10take the terms two at a time(xy + 2x) + (5y + 10)

Factor xy + 2x + 5y + 10

xy + 6x + 7y + 42take the terms two at a time(xy + 6x) + (7y + 42)

Factor xy + 6x + 7y + 42

Expanding Two Brackets

Four products, one from each pair of terms.

Every term in one bracket multiplies every term in the other, making four pieces

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x3x2(x + 2)(x + 3)

Draw a rectangle x + 3 wide and x + 2 tall. Its whole area is that product.

x²3x2x6x3x2(x + 2)(x + 3)

Both cuts split it into 4 pieces: x × x, x × 3, 2 × x, 2 × 3.

x²3x2x6x3x2x² + 5x + 6

The two middle pieces are both lots of x, so 3x and 2x join into 5x.

x²3x2x6x3x2x² + 6 misses two pieces

Take only the corner pieces and 5x is left out — that is the usual slip.

Now you

x6x6

Expand (x + 6)(x + 6)

x−1x3

Expand (x + 3)(x − 1)

Factoring Quadratics

Find the pair that multiply and add right.

Factoring a quadratic means finding the pair that multiply and add correctly

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x²x?x?x² + 5x + 6

x² + 5x + 6 came from a rectangle. Which split of its sides makes 5x and 6?

x²3x2x6x3

2 × 3 = 6 and 2 + 3 = 5, so 2 and 3 are the numbers.

Now you

a pair multiplying to −5, adding to 4

Factor x² + 4x − 5

a pair multiplying to −6, adding to 5

Factor x² + 5x − 6

Factoring Quadratics with a Leading Coefficient

The pair hunt moves to a times c.

A leading coefficient moves the pair hunt to a times c, then the middle term splits

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the 2 in front changes the pair hunt

A 2 sits in front now, so the pair must multiply to 2 × 3 = 6, not to 3 alone.

ac = 2 × 3 = 6the pair: 1 and 61 × 6 = 6 and 1 + 6 = 7

Hunt the pair for 6 that adds to 7: 1 and 6. Those split the middle term.

split 7x with the pair

Split 7x as x + 6x. Nothing changed — the two pieces still add back to 7x.

factor each half
x(2x + 1) + 3(2x + 1)the shared bracket comes out(2x + 1)(x + 3)

Factor each half and the same bracket appears twice — pull it out, and done.

2x²6xx3x32x12x² + 7x + 3

The rectangle checks it: the four pieces add back to 2x² + 7x + 3.

Now you

2x²2x3x3????2x² + 5x + 3

Factor 2x² + 5x + 3

5x²5x2x2????5x² + 7x + 2

Factor 5x² + 7x + 2

Algebraic Identities

The expansions that keep coming back.

A few expansions come up so often they are worth knowing by sight

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a²ababb²ababa² + 2ab + b²

A square of side a + b holds a², two ab strips and b² — so the middle is 2ab.

(x−b)²b(x−b)b(x−b)b²x − bbx − bbthe three pieces left: x² − b²

Each piece left keeps an (x − b) side, so the three together are (x − b)(x + b).

Now you

x²7x7x49x7x7

Expand (x + 7)²

(x−1)²1²x − 11x − 11x² − 1

Factor x² − 1

The Sum and Difference of Two Cubes

A linear bracket times a quadratic one.

A sum or difference of two cubes factors into a linear bracket times a quadratic one

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every term of the first bracket times every term of the second
the four middle terms cancel in pairs

Multiply it out and the middle terms cancel, leaving only the two cubes.

change the sign in each bracket

Flip the sign in the linear bracket and the middle sign in the quadratic one for the sum.

8 is 2 cubed
a = x and b = 2

Write the number as a cube first, then a and b can be read straight off.

27 is 3 cubed
the middle sign is the opposite of the bracket sign

For a sum: 27 is 3³, so x³ + 27 factors as (x + 3)(x² − 3x + 9).

Now you

27 is 3 cubed

Factor x³ − 27

125 is 5 cubed

Factor x³ − 125

Finding Unknown Coefficients by Matching

Two names for one expression fix every letter.

Two expressions equal for every x must agree part by part, which pins the unknowns

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expand the left side

Expand the left side. One expression is now written out in two ways.

the x terms must agree
a + b = 7and the plain numbers must agreeab = 12

Equal for every x is strong: each matching part must be equal on its own.

a + b = 7, ab = 12the pair that adds to 7 and multiplies to 12a = 3, b = 4

One statement became two, and the pair adding to 7 and timesing to 12 is 3 and 4.

p(x + 3) = 4x + qexpand the left sidepx + 3p = 4x + qmatch the x termsp = 4q = 3 × 4q = 12

Same move on p(x + 3) = 4x + q: the x terms give p, and p then gives q.

Now you

cx + 30
expand the left side
cx + 30

(x + 6)(x + 5) is x² + cx + 30. What is c?

cx + 18
expand the left side
cx + 18

(x + 3)(x + 6) is x² + cx + 18. What is c?

Algebraic Fractions

Cancel only a factor of the whole thing.

An algebraic fraction cancels only what is a factor of the whole top and bottom

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3x / 6x

3x over 6x: the x is a factor of both, so it goes.

write both as factors
the shared 3 and x cancel
1 ÷ 2

What is left is one half, and only a whole factor could be canceled.

Now you

2x/6x

Simplify 2x / 6x

3x/6x

Simplify 3x / 6x

Adding and Multiplying Algebraic Fractions

Common bottoms to add, straight across to multiply.

Algebraic fractions add over a common denominator and multiply straight across

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sixths fit halves and thirds

Same rule as unlike fractions: rewrite both over 6, and then the tops can add.

x/2 = 3x/6x/3 = 2x/6

Cut x into sixths: half of it is three sixths, a third is two — five in all.

tops multiply, bottoms multiply

Multiplying is easier: multiply the numerators, then the denominators — x²/6.

bottoms added — the usual slip

The slip to dodge: adding the bottoms. No fifths ever appear — the parts are sixths.

Now you

straight across, top and bottom

What is (x/5) × (x/2)?

x/3x/2

What is x/3 + x/2?

Recognizing Equivalent Expressions

Equal at every x, or a near-miss that is not.

Equivalent means equal at every x — one substitution can refute, none can prove

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3(x + 2) = 3 × x + 3 × 2every x agrees3(x + 2) = 3x + 6

Expanding never changes the value: 3(x + 2) and 3x + 6 agree at every x.

3(x + 2) = 3x + 6the 2 missed its × 3

3x + 2 is the near-miss — the 2 never met the × 3. Try x = 1: 9 against 5.

the middle term is real

Squaring a sum grows a middle term: (x + 3)² is x² + 6x + 9, never x² + 9.

and 2x: equal at x = 2
one match is not equivalence
x = 3: 9 against 6

Agreeing at one x proves nothing: x² and 2x meet at x = 2. Algebra covers every x.

Now you

a square of a sum has three terms

Which is equivalent to (x + 5)²?

2(x + 3)the outside multiplies everything inside

Which is equivalent to 2(x + 3)?

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