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Product, quotient and power rules, zero, negative and rational exponents, and adding, subtracting and multiplying polynomials.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you learn where the exponent rules come from — counting copies of the base — so that $x^a x^b$, $\frac{x^a}{x^b}$, $(x^a)^b$, $x^0$, $x^{-n}$ and $x^{m/n}$ stop being six things to remember and become one. Then you use them on whole polynomials: adding and subtracting by collecting like terms, and multiplying so that every term meets every term.
You already know that $x^3$ means $x \times x \times x$, and that $2^5 = 32$. You already add and subtract whole numbers, including negative ones. Every rule in this lesson is one of those two facts: count the copies, then do the arithmetic on the count.
Base and exponent: in $x^5$, $x$ is the base and $5$ is the exponent.
Reciprocal: $\frac{1}{b}$. A negative exponent means a reciprocal: $b^{-n} = \frac{1}{b^n}$.
Radical: a root. $\sqrt[3]{8} = 2$, and the same thing is written $8^{1/3}$.
Rational exponent: an exponent written as a fraction. $b^{m/n}$ is the $n$th root of $b$, raised to the power $m$.
Every rule comes from counting copies. $x^a \cdot x^b$ is $a$ copies of $x$ next to $b$ more, so it is $x^{a+b}$. In $\frac{x^a}{x^b}$ the copies underneath cancel copies on top, leaving $x^{a-b}$. $(x^a)^b$ is $b$ lots of $a$ copies, so $x^{ab}$. The subtraction rule is what forces the last two: $\frac{x^3}{x^3} = x^0$ and also equals $1$, so $x^0 = 1$; and $\frac{x^2}{x^5} = x^{-3}$ and also equals $\frac{1}{x^3}$, so $x^{-n} = \frac{1}{x^n}$. A fractional exponent follows the same way: $b^{1/2} \cdot b^{1/2} = b^1$, so $b^{1/2}$ must be $\sqrt{b}$, and $b^{m/n}$ is the $n$th root raised to the power $m$.
Another way: diagram
A row of small $x$ tiles: three tiles bracketed as $x^3$, five more as $x^5$, and the whole row of eight bracketed underneath as $x^8$.
Another way: story
Do not memorise five rules; memorise one question. 'How many copies of the base am I left with?' Multiplying gains copies, dividing loses them, and a bracket repeats whatever is inside it.
"$x^3 \cdot x^5 = x^{15}$." Multiplying powers adds the exponents: $x^8$. Multiplying the exponents is the rule for a power of a power.
"$b^{-2}$ is negative." It is $\frac{1}{b^2}$, which is positive and smaller than $1$. The minus sign flips the number over; it does not move it below zero.
"$(2x^2)^3 = 2x^6$." The bracket raises everything inside, the $2$ included: $2^3 x^6 = 8x^6$.
The bracket raises both parts: $3^3$ and $(x^2)^3$.
Everything inside is cubed.
$3^3 = 27$ and $(x^2)^3 = x^6$.
A power of a power multiplies.
So $(3x^2)^3 = 27x^6$.
The denominator $3$ is the root: $\sqrt[3]{8} = 2$.
Take the root first to keep the numbers small.
The numerator $2$ is the power: $2^2 = 4$, so $8^{2/3} = 4$.
The bracket first: $(2x^3)^2 = 4x^6$.
Now divide: $\frac{4x^6}{x^4} = 4x^{6-4} = 4x^2$.
Simplify $x^{5} \cdot x^{7}$ as a single power.
Answer:
Simplify $\dfrac{y^{11}}{y^{6}}$ as a single power.
Answer:
Simplify $(3x^{3})^{3}$.
Answer:
Evaluate $2^0 + 2^{-2}$. Give a fraction or a decimal.
answer
You already collect like terms — $5x + 3x = 8x$ — and you already expand a single bracket, $3(x + 2) = 3x + 6$. Adding polynomials is the first of those, and multiplying them is the second done several times. The only genuinely new habit is being careful with a minus sign in front of a bracket.
Polynomial: a sum of terms $c x^k$ with whole-number powers, such as $4x^3 - x + 7$.
Like terms: terms with exactly the same letter and power. $3x^2$ and $-5x^2$ are like; $3x^2$ and $3x$ are not.
Degree: the highest power in the polynomial. $4x^3 - x^5 + 7$ has degree $5$.
Binomial: a polynomial with two terms, such as $x + 3$.
To add or subtract, line up like terms and do the arithmetic within each power; nothing else changes. A subtraction is safest done in two moves: distribute the minus sign over the whole bracket first, then add. To multiply, every term of the first factor multiplies every term of the second, and then like terms are collected: $(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$. Two products come up so often they are worth recognising rather than recomputing — $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)(a + b) = a^2 - b^2$.
Another way: diagram
A two-by-two grid for $(x + 3)(x - 5)$: the cells hold $x^2$, $-5x$, $3x$ and $-15$, with the two middle cells shaded to show they are like terms that add to $-2x$.
Another way: story
Multiplying brackets is the area of a rectangle whose sides have been cut into pieces. Every piece of one side meets every piece of the other, which is exactly why every term meets every term.
"$(5x^2 - x) - (2x^2 + 3x - 7)$ is $3x^2 - x - 7$." The minus sign in front flips every term of the second bracket: $-2x^2 - 3x + 7$. So the answer is $3x^2 - 4x + 7$.
"$(x + 3)^2 = x^2 + 9$." Squaring a bracket is not squaring each term. $(x + 3)^2 = (x + 3)(x + 3) = x^2 + 6x + 9$; the middle term is the one people lose.
"$3x^2 + 2x = 5x^3$." They are not like terms and they do not combine at all. Adding is only allowed within a power.
Distribute the minus: $5x^2 - x - 2x^2 - 3x + 7$.
Every term of the second bracket flips sign.
Collect: $5x^2 - 2x^2 = 3x^2$ and $-x - 3x = -4x$.
Like terms only.
The answer is $3x^2 - 4x + 7$.
$(2x - 1)^2$ means $(2x - 1)(2x - 1)$.
A square is the bracket times itself.
Every term meets every term: $4x^2 - 2x - 2x + 1$.
Four products.
Collect the middle: $4x^2 - 4x + 1$.
$3x \times 2x^2 = 6x^3$ and $3x \times (-x) = -3x^2$.
$3x \times 4 = 12x$, so the answer is $6x^3 - 3x^2 + 12x$.
Add: $(x^2 - 8x - 7) + (x^2 + 3x + 7)$.
Answer:
Subtract: $(4x^2 - 4x) - (3x^2 - 5x - 8)$.
Answer:
Expand $(x - 7)(x + 5)$.
Answer:
Expand $5x(4x^2 - x - 1)$.
Answer:
A rectangular flower bed is $6$ m by $12$ m. A path of the same width $x$ metres is laid all the way around it. Write the area of the bed **and** path together, expanded.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Evaluate $64^{2/3}$.
answer
What is the degree of $5x^{2} - x^{6} + 4$?
You can apply the exponent rules and compute with polynomials. Explain why $x^0 = 1$ follows from the division rule, and expand $(2x - 1)^2$ without losing the middle term.
8. Your turn: simplify $\dfrac{(2x^3)^2}{x^4}$, step 2
19. Your turn: expand $3x(2x^2 - x + 4)$, step 2