Simplifying, Expanding and Factoring Expressions

An expression is a way to write a calculation. It can use letters for numbers, as in $5a-2b+8$. A letter is a variable. Its value can change, or it can stand for an unknown number.

Read one term at a time

The parts joined by addition or subtraction are terms. Keep the sign before each term. The expression above has three terms: $5a$, $-2b$, and $8$. You can read it as $5a+(-2b)+8$.

Inside a term, multiplication holds the parts together. The term $5a$ means $5\cdot a$. It does not mean $5+a$ or a two-digit number. If $a=3$, then $5a=15$.

A coefficient is the number multiplying the variable part. In $-2b$, the coefficient is $-2$. The minus sign matters. A constant term has no variable; here it is $8$.

Some coefficients are hidden. The term $b$ means $1b$, so its coefficient is $1$. The term $-b$ means $-1b$, so its coefficient is $-1$. A coefficient can also be a fraction or decimal. Both $\frac{a}{4}$ and $\frac{1}{4}a$ mean one quarter of $a$.

Write products in a shorter way

Put the coefficient first, followed by the letters. Write letters in alphabetical order. For example:

Repeated letters become powers: $a\cdot a=a^2$ and $b\cdot b\cdot b=b^3$. The small raised number counts repeated factors. A fraction bar shows division, so $a÷5=\frac{a}{5}$.

The letters next to each other still mean multiplication. The term $6ab$ is one term, even though it contains two letters.

Name the parts of an expression

Consider $9m-n+0.2mn-6$.

The remaining term, $-6$, is the constant term. There are four terms altogether. The product $mn$ stays inside one term; it is not two terms joined by addition.

After like terms have been combined, expressions also have names based on their number of terms. A monomial has one term, such as $7ab$. A binomial has two unlike terms, such as $3a+2$. A trinomial has three unlike terms, such as $a^2+3a-2$.

An expression has no equals sign. Writing $3a+2=14$ makes an equation: it states that two expressions are equal.

You can add five lengths of rope to three equal lengths. Together they make eight lengths. Algebra uses the same idea: $5x+3x=8x$. The number of copies changes, but the size of each copy stays $x$.

Decide which terms match

Like terms have the same variables with the same powers. Their coefficients can differ. Thus $7a$ and $-2a$ match, as do $3x^2$ and $x^2$. Also, $ab$ and $ba$ match, since changing the order of multiplication leaves the product unchanged.

The terms $a$ and $a^2$ are unlike. So are $a$ and $b$, or $ab$ and $a^{2}b$. You cannot combine $4a+3b$ into one term. All constant terms are like terms with each other.

Group the terms, keeping their signs

Simplify $8a+3b-5a+9-b-4$.

1. Group the $a$-terms: $8a-5a=3a$.
2. Group the $b$-terms: $3b-b=2b$, because $b$ means $1b$.
3. Combine the constants: $9-4=5$.

The result is $3a+2b+5$. The coefficient rule is the distributive property in reverse: $8a-5a=(8-5)a$.

Terms can move when you add them, but their signs move too. In $9-6x+2x$, the coefficients are $-6$ and $2$. Their sum is $-4$, giving $9-4x$.

Negative totals work the same way: $-2n-7n=-9n$. Opposite terms cancel because their coefficients add to zero: $6p-6p=0$.

Keep different powers separate

For $5x^2+2x-3x^2+7x$, collect two groups. The squared terms give $2x^2$; the first-power terms give $9x$. The answer is $2x^2+9x$.

Adding does not create a new power. The sum $x+x$ is $2x$. The expression $x^2$ means multiplication, $x\cdot x$. With $x=3$, the sum is $6$ and the product is $9$.

Work with fractions and decimals

The variable part still stays unchanged. Add the numbers using a common denominator:

For decimals, $1.4p+0.3p-0.9p=0.8p$. An unwritten coefficient still counts: $n-\frac{1}{4}n=\frac{3}{4}n$.

Know when to stop

In simplest form, parentheses are expanded, products are compact, and no like terms remain. For example, $2(x+4)+3x$ first expands to $2x+8+3x$, then becomes $5x+8$. A request to factor asks for a different form, with a factor outside parentheses.

You can check the arithmetic by substitution. At $a=2$ and $b=3$, the longer expression above gives $16+9-10+9-3-4=17$. Its shorter form gives $6+6+5=17$.

Two expressions are equivalent when they have the same value for every value of their variables. For example, $6x+x$ and $7x$ are equivalent. This is stronger than matching at just one input.

Explain why a rewrite works

The properties of operations let you change an expression's form without changing its value.

For $9a+4-2a+1$, rewrite subtraction as adding a negative. You can then reorder the additions:

Group the like terms and add their coefficients. This gives $7a+5$. Each step works for every value of $a$.

Subtraction and division cannot be reordered this way. At $a=8$, $a-3$ gives $5$, while $3-a$ gives $-5$. Moving a term with its sign is different from reversing the order of a subtraction.

Use numbers to catch a wrong claim

Suppose someone says $4(x+2)$ equals $4x+2$. Choose $x=3$. The first expression gives $4(3+2)=20$. The second gives $4\cdot 3+2=14$. That one mismatch is enough to show they are not equivalent.

A match at one input is weaker evidence. Both $x^2$ and $3x$ give $9$ at $x=3$. At $x=2$, they give $4$ and $6$.

Try inputs such as $2$, $-3$, or a fraction when checking work. Zero and one can hide errors: multiplying by zero removes many differences, and different positive powers of one all equal one.

Tables, graphs, and proof

A table compares several inputs at once. For $3(x+2)$ and $3x+6$, inputs $0$, $1$, and $2$ give $6$, $9$, and $12$ in both columns. The table supports the claim. The algebraic reason is distribution: $3(x+2)=3x+6$ for every input.

Equivalent expressions also produce the same graph. Plotting $y=3(x+2)$ and $y=3x+6$ gives the same line.

Expression, equation, identity, or formula?

An expression, such as $2x+5$, has no equals sign. An equation states equality, as in $2x+5=13$. This equation is true at $x=4$. An identity is true for every input, such as $2(x+5)=2x+10$. Some books mark identities with $\equiv$. A formula calculates one quantity from others, such as $P=2l+2w$ for a rectangle's perimeter.

A product asks you to multiply. In $2a\cdot 7b$, both the numbers and letters are factors. You may reorder and regroup those factors, which helps you work with the numbers first.

Multiply the two parts

Multiply the coefficients, then collect the letters. Unlike terms can be multiplied: $2a\cdot 7b$ becomes one term. Their sum, $2a+7b$, stays as two terms.

Signs follow ordinary multiplication rules. One negative factor and one positive factor give a negative product. Two negative factors give a positive product:

Count repeated factors

A power is a short way to write repeated multiplication. In $x^3$, the base is $x$ and the exponent is $3$. It means $x\cdot x\cdot x$. The powers $x^2$ and $x^3$ are read as “x squared” and “x cubed.”

For $4x\cdot 3x$, the coefficients give $12$. There are two factors of $x$, so the result is $12x^2$. Writing $12x$ loses a factor.

When multiplying powers of the same base, add their exponents. The product $x^2\cdot x^3$ has five factors of $x$, giving $x^5$. In $3x^2\cdot 5x^3=15x^5$, the numbers multiply while the exponents add.

Different letters are counted separately. For example:

The coefficient is $2\cdot 3=6$. There are three $a$ factors and three $b$ factors. Write the letters in alphabetical order.

Fractions, decimals, and several factors

For $\frac{3}{5}a\cdot 10b$, the numbers give $\frac{3}{5}\cdot 10=6$, leaving $6ab$. Similarly, $0.2y\cdot 3y=0.6y^2$.

With three factors, $2p\cdot (-3q)\cdot p$ has coefficient $-6$, two factors of $p$, and one of $q$. The product is $-6p^{2}q$.

Check what an exponent belongs to

In $4x^2$, only $x$ is squared. In $(4x{)}^2$, the whole term is repeated:

At $x=3$, $4x^2$ gives $4\cdot 9=36$. The grouped square gives ${12}^2=144$. Parentheses make the difference.

Adding and multiplying coefficients are also different. The sum $4x+3x$ is $7x$. The product $4x\cdot 3x$ is $12x^2$. Use the operation sign to decide which rule applies.

Division can undo a multiplication. If three equal groups make $21x$, each group is $7x$. Dividing the coefficient by three leaves the letter factor unchanged: $21x÷3=7x$.

Divide by a number

Work on the coefficient and keep the variable part. For example, $\frac{-28ab}{7}=-4ab$.

When the coefficient is not a whole number, keep a simplified fraction. In $\frac{10x}{15}$, divide the top and bottom numbers by five to get $\frac{2x}{3}$. This also means $\frac{2}{3}x$.

Multiplying by a fraction can describe the same calculation. One fifth of $20y$ is $\frac{1}{5}\cdot 20y=4y$, which is the same as $20y÷5$.

Divide by another term

Write out the factors to see what cancels:

A common nonzero factor divided by itself gives one. Both a numerical factor of two and a letter factor of $b$ cancel. The factor $a$ stays in the numerator.

The divisor cannot be zero. In this example, $b\mathrm{\ne }0$. The shorter expression $7a$ can be calculated at $b=0$, but the original quotient cannot.

Count the factors left over

For $\frac{15x^3}{5x}$, the coefficients give $3$. One factor of $x$ cancels from the three in the numerator, leaving two:

This explains the exponent shortcut. When dividing powers of the same nonzero base, subtract the denominator's exponent from the numerator's. Thus $\frac{x^5}{x^2}=x^3$ for $x\mathrm{\ne }0$.

Check division by multiplying back. For the quotient above, $3x^2\cdot 5x=15x^3$. A proposed quotient of $3x^3$ would give an extra factor of $x$.

Some divisions leave no letters: $\frac{15m}{20m}=\frac{3}{4}$ for $m\mathrm{\ne }0$. All the letter factors cancel, but the numerical fraction remains.

Cancel factors, not added terms

A fraction bar divides the whole numerator. In $\frac{6x+12}{3}$, each term is divided by three:

Dividing only the constant would give $6x+4$, which leaves the first term three times too large. At $x=2$, the original is $24÷3=8$. The correct shorter form gives $4+4=8$; the incorrect form gives $16$.

The expression $4(x+3)$ means four groups of $x+3$. Each group contains one $x$ and three more. Altogether there are four copies of $x$ and twelve more: $4x+12$.

Why distribution works

The distributive property says:

The outside factor multiplies every term inside. With subtraction, $a(b-c)=ab-ac$. Expanding uses this property to write a product as a sum or difference.

A split rectangle shows the same idea. Give it height four and length $x+3$. Split the length into $x$ and three. The two part-areas are $4x$ and $12$, so the total is $4(x+3)=4x+12$.

This also helps with mental arithmetic. For example, $7\cdot 19=7(20-1)=140-7=133$.

Expand a difference

For $5(2n-3)$, multiply five by both signed terms:

Put the results together: $5(2n-3)=10n-15$. With three terms, $3(2a-b+1)=6a-3b+3$.

A multiplier can come after the parentheses too. The expression $(x+2)\cdot 6$ expands to $6x+12$.

Keep a negative multiplier together

In $-4(y-3)$, multiply by $-4$ each time. The first product is $-4y$. The second is $(-4)(-3)=12$, so the expansion is $-4y+12$.

A minus sign by itself means multiplication by $-1$. Thus $-(2x+5)=-2x-5$. Both terms change sign.

If you write $-4y-12$ for the first example, you have treated a negative times a negative as negative. At $y=0$, the original gives $-4(-3)=12$, which catches the sign error.

Fractions and decimals still multiply every term

The two products are $\frac{3}{4}\cdot 8x=6x$ and $\frac{3}{4}\cdot (-12)=-9$. Similarly, $0.2(15+5m)=3+m$.

An added number stays separate

For $6+3(x-2)$, expand the product before combining terms:

The added six is not part of the multiplier. Multiplying only the first inside term is another common error: $3(x-2)$ needs both $3x$ and $-6$. At $x=4$, the product is $3\cdot 2=6$, which agrees with $12-6$.

The outside multiplier can contain a letter. The distributive property still reaches every term inside. The extra letter becomes a factor in each product.

Let the letter reach both terms

For $x(x+4)$, make two products:

So $x(x+4)=x^2+4x$. The constant four becomes $4x$ because it is multiplied by $x$ too.

Picture a rectangle with width $x$ and length $x+4$. Splitting its length gives an $x$-by-$x$ square and an $x$-by-four rectangle. Their areas are $x^2$ and $4x$. At $x=3$, the whole rectangle is three by seven, with area $21$. The parts give $9+12=21$.

Multiply numbers and letters in each product

Expand $2a(3a-5b)$. First multiply $2a$ by $3a$. The coefficients give six, and the repeated $a$ factors give $a^2$. Next multiply $2a$ by $-5b$. This gives $-10ab$.

These terms are unlike: one has $a^2$ and the other has $ab$. Leave them separate. Having a letter in common is not enough to make terms like terms.

The same steps handle a power inside:

The first product contains three factors of $y$. The second contains just one. You cannot add their coefficients to make a single term.

Watch the sign of the whole multiplier

In $-2m(m-6)$, the multiplier is $-2m$. Its products are $-2m^2$ and $12m$, giving:

The positive second term comes from two negatives multiplying. For $-p(5-p)$, the expansion is $-5p+p^2$. Writing the highest power first gives $p^2-5p$; both orders mean the same thing.

Check for a missing factor

A frequent error in $x(x+4)$ is writing $x^2+4$. The second term is missing its factor of $x$. At $x=3$, that wrong form gives $13$, while the original gives $21$.

Another error is treating $x\cdot x$ as addition. It is $x^2$, not $2x$.

A longer expression can contain several products. Expand each product first. Then look for like terms across the whole expression.

Expand separately, then combine

Simplify $2(x+7)+3(x-4)$.

The first product gives $2x+14$. The second gives $3x-12$. Write both before combining:

At $x=2$, the original gives $2(9)+3(-2)=12$. The result gives $5\cdot 2+2=12$.

Subtract every term in a group

Adding expressions leaves their signs unchanged:

Subtraction changes both signs in the second group. Treat the minus as a multiplier of $-1$:

The last term is positive eight because subtracting negative eight adds eight. Changing only the first sign would produce $5x-3$. At $x=0$, the original is $5-(-8)=13$, which reveals that error.

Keep the full signed multiplier

In $5(2x-3)-2(x-6)$, the second multiplier is $-2$. Its expansion is $-2x+12$, not $-2x-12$. Now write the complete line:

A useful way to organize the work is to underline each multiplier together with its sign before expanding. This keeps the subtraction attached to the group it acts on.

An added or subtracted number stays separate from a product. For $9-3(2-x)$, expansion gives $9-6+3x=3x+3$. You cannot begin by calculating $9-3$, because the three belongs to a multiplication.

Combine the new terms carefully

With a variable multiplier, $x(x+6)-2(x+1)$ becomes $x^2+6x-2x-2$. Combine the first-power terms to get $x^2+4x-2$. The squared term stays separate.

Sometimes new products cancel. In $a(b+3)-b(a-2)$, expansion gives $ab+3a-ab+2b$. The opposite $ab$ terms add to zero, leaving $3a+2b$.

Fractions follow the same order. For $\frac{1}{2}(8x+6)-\frac{1}{3}(3x-12)$, the first part is $4x+3$ and the second signed part is $-x+4$. Combining gives $3x+7$.

A factor multiplies another factor to make a term. For example, $15x=3\cdot 5x$, so three and $5x$ are factors of $15x$. A common factor occurs in every term you are comparing.

Start with the coefficients

The greatest common factor, or GCF, of whole-number coefficients is the greatest positive whole number that divides them all exactly.

For $20$ and $30$, you could list factors. Their common factors are $1$, $2$, $5$, and $10$, so their GCF is $10$.

Prime factors offer another method:

They share one factor of two and one factor of five. Multiplying these gives $10$. Taking just two would find a common factor, but not the greatest one.

Include letters shared by every term

For $20x^{2}y$ and $30x{y}^3$, the coefficients give ten. Each term has at least one factor of $x$ and at least one factor of $y$. The GCF is therefore $10xy$.

Use the smallest power of each shared letter. Only one $x$ can be taken from the second term, even though the first has two. Only one $y$ can be taken from the first term, even though the second has three.

Check by division:

This describes the factors written in the terms. It does not depend on choosing numerical values for the letters.

A missing letter cannot be common

For $14a$ and $21$, the GCF is seven. The second term has no factor of $a$, so $7a$ is not a common factor.

For $8m^{2}n$ and $12mn$, both letters are shared. Their smallest powers are one, so the GCF is $4mn$.

With three terms, check all three. The terms $12p$, $18pq$, and $24p^2$ share $6p$. The letter $q$ appears in only one term, so leave it out of the GCF.

If the only common factor is one, there is no larger common factor to remove. For example, $5x+7$ has no shared letter, and the coefficients have GCF one.

Check whether you have finished

After removing a proposed common factor, look at what remains. Dividing $20x^{2}y$ and $30x{y}^3$ by $5xy$ leaves $4x$ and $6y^2$. They still share two, so $5xy$ was too small. Using $10xy$ leaves $2x$ and $3y^2$, which share no further numerical or variable factor except one.

Factoring reverses expanding. Expanding $7(x+2)$ gives $7x+14$. Factoring $7x+14$ finds the shared seven and writes it outside parentheses.

Divide every term by the common factor

Factor $16x+24$ completely.

1. Find the GCF of the coefficients: eight.
2. Put eight outside parentheses.
3. Divide each term by eight: $16x÷8=2x$ and $24÷8=3$.

The factored form is $8(2x+3)$. Expanding checks it: $8\cdot 2x+8\cdot 3=16x+24$.

Writing $8(2x+24)$ divides only the first term. Expansion would make the constant $192$, so it cannot be equal to the original expression.

Divide the letter factors too

For $10x^2+15x$, the GCF is $5x$. Dividing gives $2x$ and three, so:

One factor of $x$ remains in the first inside term. None remains in the second. Similarly, $6ab+9a=3a(2b+3)$.

Keep a one when a whole term is removed

In $7y^2-7y$, the GCF is $7y$. The first quotient is $y$; the second is $-1$. Thus $7y^2-7y=7y(y-1)$.

A term divided by itself gives one, not zero. Losing the one would lose the entire original term when you expand again.

Three-term expressions follow the same method. For $12a-18b+6$, divide all three terms by six:

Take out the whole GCF

The form $2(8x+12)$ equals $16x+24$, but it is not completely factored. The inside terms still share four. Taking out the GCF at the start gives $8(2x+3)$, with no common factor left inside except one.

“Factor completely” asks for that complete common-factor form. “Expand and simplify” asks for the terms written out instead.

Negative and fractional factors

A negative factor can make the first inside term positive. For $-15n+20$, take out $-5$:

The inside negative four gives the positive twenty on expansion. The form $5(-3n+4)$ is equivalent too.

You can also take out a fractional or decimal factor. For example:

In the first example, $2÷\frac{1}{4}=8$. In the second, $-1.2÷0.4=-3$. Expanding multiplies each quotient by the outside factor to recover the original terms.

The perimeter is the distance around a shape's boundary. If side lengths contain letters, add those expressions just as you would add numerical lengths.

Add every side

A triangle has sides $2x+1$, $x+4$, and $2x+3$ centimeters. Its perimeter is:

At $x=2$, the sides are five, six, and seven centimeters. Their total is 18 centimeters, which agrees with $5\cdot 2+8=18$.

Side lengths must be positive and must describe a possible shape. Here $x=2$ meets those requirements.

Count both pairs of rectangle sides

A rectangle has two lengths and two widths. With length $l$ and width $w$, its perimeter is $P=2l+2w$. Grouping one length and one width also gives $P=2(l+w)$.

For length $2x+3$ and width $x+1$ centimeters:

Expand each product: $4x+6+2x+2$. Then combine like terms to get $6x+8$ centimeters.

At $x=3$, the rectangle is nine by four centimeters. Its perimeter is $2(9+4)=26$ centimeters. The expression gives $6\cdot 3+8=26$ too.

Adding just the two different side labels counts only half the boundary. Multiplying them would calculate area instead.

Equal sides can be counted in groups

A regular polygon has all sides equal. If there are $n$ sides of length $s$, its perimeter is $ns$. A regular pentagon with side $3y+2$ centimeters has perimeter:

The five multiplies both parts of the side length. It counts five complete sides.

Follow the outside of a composite shape

A composite shape joins or removes simpler shapes. Trace the whole boundary, including the edges of any indentation. A line shared by two joined pieces is inside the shape, so it does not count toward the perimeter.

Suppose a rectangle is 12 centimeters long and seven centimeters wide. Remove a three-by-two-centimeter rectangle from one corner. The cut removes outside lengths of three and two centimeters. It creates new boundary edges of two and three centimeters. The total perimeter stays $2(12+7)=38$ centimeters.

If a missing edge lies along the same direction as known edges, use their total lengths. For example, two horizontal pieces that together span 12 centimeters must add to 12 centimeters. This helps you write an expression before collecting like terms.

Area measures how much surface a shape covers. A rectangle's area is length times width. Keep a whole side length in parentheses when it contains more than one term.

A rectangle as one product or two parts

A rectangle is five centimeters wide and $2x+3$ centimeters long. Its area is $5(2x+3)$ square centimeters. Expanding gives $10x+15$.

Splitting the length into $2x$ and three gives two rectangles. Their areas are $10x$ and 15 square centimeters. Adding the parts gives the same total as multiplying the full dimensions.

The expression $5\cdot 2x+3$ would add only three square centimeters after the first product. It misses the strip's full area of 15 square centimeters.

A letter can be a side length

For sides $x$ and $x+6$ centimeters, the area is:

The parts are a square of area $x^2$ and a rectangle of area $6x$. At $x=2$, the shape is two by eight centimeters, with area 16 square centimeters. The expanded form gives $4+12=16$.

A square with side $4a$ has area $(4a{)}^2=16a^2$. Both copies of the side include the four. Its perimeter is $4\cdot 4a=16a$, which measures a different quantity.

Use the right height

A parallelogram has area base times perpendicular height. For base $x+4$ centimeters and perpendicular height three centimeters, the area is $3(x+4)=3x+12$ square centimeters. A sloping side is not the perpendicular height.

A triangle has half the area of a parallelogram with the same base and height. For base $6x$ centimeters and perpendicular height five centimeters:

First, half of $6x$ is $3x$. Multiplying by five gives $15x$ square centimeters.

Add parts or subtract a cutout

For a composite shape, split the surface into rectangles that do not overlap. Add their areas. Alternatively, find the larger rectangle's area and subtract a missing piece.

A rectangle $4x$ centimeters long and seven centimeters wide loses an $x$-by-three-centimeter piece. Its remaining area is $28x-3x=25x$ square centimeters.

Factoring can recover a rectangle's dimensions. If its area is $8x+20$ square centimeters and its width is four centimeters, write $8x+20=4(2x+5)$. The other side is $2x+5$ centimeters.

Area uses square units, such as ${\text{cm}}^2$. Perimeter uses length units, such as centimeters. The units help you check whether your calculation describes the quantity asked for.

An expression should match what happens in the situation. First decide what each letter means. Then choose the operations and group anything that is repeated as a whole.

Read the order in the words

“Six less than $n$” means $n-6$. Start with the number, then remove six. “Triple the sum of $x$ and two” means $3(x+2)$. The whole sum is tripled. “Two more than triple $x$” means $3x+2$.

For “a number increased by four, then halved,” write $\frac{1}{2}(n+4)$. Parentheses preserve the order: add first, then take half.

Count repeated costs

A club buys five notebooks at $n$ dollars each and two pens at $p$ dollars each. It then buys another notebook. The cost is $5n+2p+n=6n+2p$ dollars.

If $n=3$ and $p=2$, the total is $18+4=22$ dollars.

For three people each buying a meal at $m$ dollars and a drink at $d$ dollars, use $3(m+d)$. This factored form shows three complete purchases. Its expanded form, $3m+3d$, shows the meal and drink totals separately.

Separate a shared fee from a repeated cost

Suppose an activity costs nine dollars per person plus one shared 15-dollar equipment fee. For $n$ people, the total is $9n+15$ dollars.

The fee is added once. The expression $n(9+15)$ would charge it to every person. At four people, the correct total is $36+15=51$ dollars. Charging the fee four times would give $4\cdot 24=96$ dollars.

Use a new form to compare prices

Plan A costs $4(x+6)$ dollars. Plan B costs $4x+19$ dollars. Expand Plan A to get $4x+24$. Both contain the same $4x$ charge, so Plan A costs five dollars more for any allowed value of $x$.

No particular value of $x$ is needed to compare those fixed parts. The shared variable term accounts for the part that changes in both plans.

Turn a percent change into one multiplier

An increase of 5% adds $0.05a$ to an original amount $a$:

The one in $1.05$ keeps the whole original amount. The $0.05$ adds the increase. Multiplying by $0.05$ alone would give only the increase.

A 20% discount leaves 80% of a price $p$, so $p-0.2p=0.8p$. For a 40-dollar price, the discount is eight dollars and the amount paid is 32 dollars.