Equivalent expressions

There's usually no story and no figure. You get an expression and four choices in the same form: all multiplied out, all factored, all one fraction, or all roots. That's the form your answer needs, so look at the choices first.

The last line might say:

• “Which expression is equivalent to $\frac{3w-2}{w+4}$, where $w\gt0$?”
• “If $p=2m^2-5$ and $q=m^2+3m-4$, which expression is equivalent to $3q-p$?”
• “Which of the following expressions is(are) a factor of $3t^2-10t-8$?”
• On a grid-inA question where you type your own answer instead of picking one of four.Read the lesson: “The expression $(x+3)(x-7)$ can be rewritten as $x^2+bx-21$, where $b$ is a constantA number that doesn’t change, like the 7 in 5x + 7, or a letter such as p that stands for one fixed value.Read the lesson. What is the value of $b$?”

Usually nothing gets solved, so don't look for a value of $x$. A condition like $w\gt0$ is there because a denominatorThe bottom number of a fraction, like the 4 in 3/4. or a root would break for some numbers.

If the expression comes in a sentence, such as a price per ticket plus a fee per order, write it out from the words first.

You can test a choice with one number. Say the expression is $2(x+6)-3x$, and two of the choices are $-x+12$ and $-x+6$.

1. Pick a number that isn't 0 or 1.

   $$x=3$$

2. Put it into the original.

   $$2(3+6)-3(3)=9$$

3. Put it into $-x+12$. Same value, so that choice stays.

   $$-3+12=9$$

4. Put it into $-x+6$. It gives 3, so that choice is out.

   $$-3+6=3$$

At $x=1$, $x$, $x^2$ and $x^3$ all equal 1, and at $x=0$ they're all 0, so different choices can tie. If two choices still match, try a second number. Stay inside any condition the question gives, so for $x\gt9$ use 10 or 12, and never use a number that makes a denominator zero. The plugging-in lesson has more on picking numbers.

On a grid-in, a number can find the constant. For $(x+3)(x-7)=x^2+bx-21$, $x=2$ gives $5(-5)=-25$ on the left and $4+2b-21$ on the right, so $2b-17=-25$ and $b=-4$. See the grid-in lesson for entering fractions and negatives.

For a single distribution, working by hand is faster than typing. Desmos basics covers typing expressions in.

When both parts have more than one term, every term in the first multiplies every term in the second. Two terms times three terms gives six products. Here's $(y-3)(2y^2-y+4)$.

1. Multiply $y$ by all three terms.

   $$y(2y^2-y+4)=2y^3-y^2+4y$$

2. Multiply $-3$ by all three terms. Minus times minus is plus.

   $$-3(2y^2-y+4)=-6y^2+3y-12$$

3. Put the six products together.

   $$2y^3-y^2-6y^2+4y+3y-12$$

4. Combine like powers and write them from the highest power down, the way the choices print them.

   $$2y^3-7y^2+7y-12$$

When the question names two expressions and combines them, put each one in parentheses first. If $p=2m^2-5$ and $q=m^2+3m-4$, here's $3q-p$.

1. Put each whole expression in parentheses.

   $$3(m^2+3m-4)-(2m^2-5)$$

2. Multiply all three terms of $q$ by 3, and change both signs in $p$.

   $$3m^2+9m-12-2m^2+5$$

3. Combine: $3m^2-2m^2=m^2$ and $-12+5=-7$.

   $$m^2+9m-7$$

Without the parentheses you'd write $3m^2+3m-4-2m^2-5$. The 3 would reach only the $m^2$, and the $-5$ would never turn into $+5$.

$12+18$ is $6(2+3)$, because 6 goes into both numbers. You can pull a factor out of terms with letters the same way.

1. Start. 4 goes into 8, 12 and 20, and every term has at least $w^2$.

   $$8w^6+12w^4-20w^2$$

2. Divide the first term by $4w^2$. $w^6$ is $w^2\cdot w^4$.

   $$8w^6\div4w^2=2w^4$$

3. Divide the second term.

   $$12w^4\div4w^2=3w^2$$

4. Divide the last term. The $w^2$ cancels.

   $$-20w^2\div4w^2=-5$$

5. Put $4w^2$ outside and the results inside.

   $$4w^2(2w^4+3w^2-5)$$

Taking out $w^4$ would leave the last term with a negative power. To check, multiply back out and make sure the original terms come back.

Try it with numbers first: $10^2-3^2=100-9=91$, and $(10-3)(10+3)=7\cdot13=91$.

$9m^2$ is $(3m)^2$ and $16n^2$ is $(4n)^2$. A number in front stays in front: $3(9m^2-16n^2)=3(3m-4n)(3m+4n)$. An even power is a square too, since $r^6=(r^3)^2$. So $r^6-25s^6=(r^3-5s^3)(r^3+5s^3)$. A sum like $9m^2+16n^2$ doesn't factor this way.

The other pattern has a middle term. You can see where it comes from in $13^2=(10+3)^2=100+60+9=169$, where the 60 is $2\cdot10\cdot3$.

Some questions hide one of these inside a subtraction: $(3r^3+2)(r^3-4)-(r^3-5)(r^3+5)$.

1. Four products for the first pair.

   $$(3r^3+2)(r^3-4)=3r^6-12r^3+2r^3-8$$

2. Combine $-12r^3+2r^3$.

   $$=3r^6-10r^3-8$$

3. One square minus another.

   $$(r^3-5)(r^3+5)=r^6-25$$

4. Subtract the whole second result, kept in parentheses.

   $$3r^6-10r^3-8-(r^6-25)$$

5. $-(-25)=+25$, and $-8+25=17$.

   $$2r^6-10r^3+17$$

$(x+8)(x-6)=x^2-6x+8x-48=x^2+2x-48$. The 2 is $8+(-6)$, and the $-48$ is $8\cdot(-6)$. Factoring runs that backward.

When the number in front of $x^2$ isn't 1, multiply it by the number on its own first. For $3t^2-10t-8$, that's $3(-8)=-24$.

1. Find two numbers that multiply to $-24$ and add to $-10$.

   $$-12\cdot2=-24,\ \ -12+2=-10$$

2. Use them to split $-10t$ into $-12t+2t$.

   $$3t^2-12t+2t-8$$

3. Pull a factor out of each pair.

   $$3t(t-4)+2(t-4)$$

4. Both parts share $t-4$, so pull it out.

   $$(3t+2)(t-4)$$

Some questions offer factors, such as I. $3t+2$ and II. $t+2$, and ask which ones work. Set each one equal to zero and put that $t$ into the whole expression. A factor makes it zero.

Only I works. If “Neither” is a choice, it can be right, so test both.

In Desmos, type it with $x$ in place of $t$, as y=3x^2-10x-8, and click where the curve crosses the horizontal axisOne of the two lines a graph is measured against, one running across and one running up.. The crossings at $x=4$ and $x=-0.667$ match the factors $t-4$ and $3t+2$, since $3t+2$ is zero at $-\frac23$.

Some questions put letters where numbers would be. Say $(3x+m)(x-4)=3x^2+bx-20$, where $m$ and $b$ are constants, and you need $b$. Multiply out the left side and match the parts.

1. Four products.

   $$(3x+m)(x-4)=3x^2-12x+mx-4m$$

2. Group the two $x$ terms.

   $$=3x^2+(m-12)x-4m$$

3. Match the numbers on their own.

   $$-4m=-20,\ \ m=5$$

4. Match the $x$ terms.

   $$b=m-12=-7$$

If a question like this asks what must be true and $b$ never gets a value, the number in front of $x^2$ and the number on its own stay fixed whatever $b$ is, but anything built from the middle term can change.

Start with 2s.

Different letters never combine, so $x^3y^4$ stays as it is.

Keep dividing by 2 and you reach zero and negative exponents: $2^2=4$, $2^1=2$, $2^0=1$, $2^{-1}=\frac12$, $2^{-2}=\frac14$.

With three letters and negative exponents, go one letter at a time. Here's the product of $u^4v^{-3}w^2$ and $u^{-6}v^5+v^2w^{-5}$.

1. Multiply by the first term.

   $$u^4v^{-3}w^2\cdot u^{-6}v^5$$

2. Add the exponents for each letter. There's no new $w$, so $w^2$ stays.

   $$=u^{4-6}v^{-3+5}w^2=u^{-2}v^2w^2$$

3. Multiply by the second term.

   $$u^4v^{-3}w^2\cdot v^2w^{-5}$$

4. There's no new $u$, so $u^4$ stays.

   $$=u^4v^{-3+2}w^{2-5}=u^4v^{-1}w^{-3}$$

5. Add the two results.

   $$u^{-2}v^2w^2+u^4v^{-1}w^{-3}$$

If the choices have no negative exponents, move those powers under a fraction bar: $\frac{v^2w^2}{u^2}+\frac{u^4}{vw^3}$. Desmos can't graph three letters at once, so check this kind by hand.

Take $8^{2/3}$. The 3 on the bottom means a cube root and the 2 on top means a square, so $8^{2/3}=\sqrt[3]{8^2}=\sqrt[3]{64}=4$.

A root with an unfamiliar number can still match, because the fraction can be scaled top and bottom. For positive $r$ and $s$, $(rs)^{5/6}=(rs)^{15/18}=\sqrt[18]{(rs)^{15}}$. Turn each choice back into a fraction exponent and reduce it. $\sqrt[6]{(rs)^{15}}$ is $(rs)^{15/6}$, which isn't $\frac56$.

A root inside a root multiplies the small numbers: $\sqrt[3]{\sqrt a}=(a^{1/2})^{1/3}=a^{1/6}$. Adding the 2 and the 3 to get $a^{1/5}$ is the usual slip.

Some questions give roots of the same base and a target. For what value of $x$ is $\frac{\sqrt[3]{(2p)^4}}{\sqrt{2p}}$ equivalent to $(2p)^{5x}$, where $p\gt0$?

1. Write each root as a fraction exponent. The power inside goes on top.

   $$\frac{(2p)^{4/3}}{(2p)^{1/2}}$$

2. Same base, so dividing subtracts the exponents.

   $$(2p)^{4/3-1/2}$$

3. $\frac86-\frac36=\frac56$.

   $$(2p)^{5/6}$$

4. Set the two exponents equal.

   $$5x=\frac56$$

5. Divide both sides by 5.

   $$x=\frac16$$

In numbers, $\frac{23}{5}=4+\frac35$, because $23=4\cdot5+3$.

When the bottom has a number in front of the letter, it divides too. Here's $\frac{12t+5}{3t-1}$, where $t\gt1$.

1. Divide the numbers in front of $t$.

   $$12\div3=4$$

2. Multiply the bottom by 4.

   $$4(3t-1)=12t-4$$

3. From $-4$, it takes 9 more to reach 5.

   $$12t+5=(12t-4)+9$$

4. Write the 4, then the leftover 9 over the bottom.

   $$4+\frac{9}{3t-1}$$

That number doesn't have to be whole. $\frac{5x+1}{2x+3}$ starts with $\frac52$.

In numbers, multiply each top by the other bottom: $\frac12-\frac13=\frac{1\cdot3-1\cdot2}{2\cdot3}=\frac16$.

1. The top, from the right side above.

   $$5n-3(n+2)$$

2. Multiply out the product. Keep it in parentheses.

   $$5n-(3n+6)$$

3. The minus changes both signs inside.

   $$5n-3n-6$$

4. Combine: $5n-3n=2n$.

   $$2n-6$$

5. Leave the bottom factored. The choices print it that way.

   $$\frac{2n-6}{n(n+2)}$$

At $n=2$, the original is $\frac54-\frac32=-\frac14$, and the answer is $\frac{-2}{8}=-\frac14$.

To check this in Desmos, type the original after f(x)= with $x$ for $n$: 5/(x+2), press the right arrow, then type -3/x. Without the arrow, everything after the first / stays in the first bottom.

In $\frac{6\cdot5}{3\cdot5}$, the 5 multiplies the whole top and the whole bottom, so it cancels and leaves 2. In $\frac{6+5}{3+5}=\frac{11}{8}$, the 5s are added, and they can't cancel.

$\frac{6(x+2)}{3(x+2)}$ does simplify to 2, for $x\neq-2$, because $x+2$ multiplies the whole top and the whole bottom.

Before each one, look at what the choices have in common. Then check your pick with a number.