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Use the square, difference-of-squares and cube identities in both directions, and add, multiply and divide 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 meet the handful of identities that do most of the algebra in this course — the square of a binomial, the difference of two squares, the sum and difference of two cubes — and you use them in both directions: left to right to expand, right to left to factor. Then you treat polynomials as things you can compute with, adding, subtracting, multiplying and dividing them exactly as you do whole numbers, so that later lessons can factor a cubic, simplify a rational expression or divide out a known root without stopping to think about the arithmetic.
You already multiply out a bracket: $3(x + 4) = 3x + 12$, because the $3$ reaches every term inside. You already know that $7 \times 7 = 49$ and that a rectangle's area is length times width. An identity is nothing more than one of those multiplications done once, in letters, and then never done again.
Identity: an equation that is true for every value of its letters. $(a + b)^2 = a^2 + 2ab + b^2$ is an identity; $x + 1 = 5$ is not — it is true only for one $x$.
Expand: turn a product into a sum of terms.
Factor: turn a sum of terms back into a product.
Binomial: an expression with two terms, like $x + 3$.
Perfect square trinomial: what a squared binomial expands to, like $x^2 + 6x + 9$.
Difference of squares: $a^2 - b^2$, which always factors as $(a + b)(a - b)$.
An identity is an equation true for every value of its letters, so you may use it in either direction: left to right to expand, right to left to factor. Four of them do most of the work in this course.
$$(a + b)^2 = a^2 + 2ab + b^2 \qquad (a - b)^2 = a^2 - 2ab + b^2$$ $$(a + b)(a - b) = a^2 - b^2$$ $$a^3 - b^3 = (a - b)(a^2 + ab + b^2) \qquad a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$
So $(x + 5)^2 = x^2 + 10x + 25$, and $x^2 - 49$ factors as $(x + 7)(x - 7)$, and $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$. Because an identity holds for every value, the letters may be numbers: $103 \times 97 = (100 + 3)(100 - 3) = 10000 - 9 = 9991$, done in your head.
Another way: picture
A square of side $a + b$ cut into four pieces: an $a$ by $a$ square, a $b$ by $b$ square, and two identical $a$ by $b$ rectangles. The whole area is $(a + b)^2$; the pieces are $a^2 + b^2 + 2ab$. The two rectangles are the term everybody forgets.
Another way: story
An identity is a multiplication somebody already did for you. You supply the two pieces, the identity supplies the answer — which is why it works just as well on $103$ and $97$ as on $x$ and $5$.
"$(a + b)^2 = a^2 + b^2$." Try $a = 3$, $b = 4$: the left side is $49$, the right side is $25$. The missing $2ab$ is the two rectangles in the picture below, and it is missing from more wrong answers than anything else in algebra.
"$a^3 - b^3 = (a - b)^3$." Try $a = 2$, $b = 1$: the left side is $7$, the right side is $1$. The cube identity has a three-term second bracket: $(a - b)(a^2 + ab + b^2)$.
Sign confusion in the cubes. In $a^3 - b^3$ the minus sits in the short bracket and every sign in the long bracket is a plus. In $a^3 + b^3$ it is the other way round: $(a + b)(a^2 - ab + b^2)$. The long bracket always disagrees with the short one in the middle.
The pattern is (first squared) + (twice the product) + (last squared).
This is $(a - b)^2$ with $a = x$ and $b = 6$.
First squared: $x^2$. Twice the product: $2 \times x \times (-6) = -12x$. Last squared: $(-6)^2 = 36$.
The middle term takes the sign of the second term.
$(x - 6)^2 = x^2 - 12x + 36$.
$27 = 3^3$, so this is $a^3 - b^3$ with $a = x$ and $b = 3$.
Recognising the cube is the whole difficulty.
The identity gives $(x - 3)(x^2 + 3x + 9)$.
Short bracket takes the minus; the long bracket is all plus.
Check the middle: expanding gives $x^3 + 3x^2 + 9x - 3x^2 - 9x - 27 = x^3 - 27$.
Everything except the ends cancels.
Write $41 = 40 + 1$ and use $(a + b)^2 = a^2 + 2ab + b^2$.
$1600 + 2 \times 40 \times 1 + 1 = 1600 + 80 + 1 = 1681$.
Expand $(x + 4)^2$.
Answer:
Use $(a + b)(a - b) = a^2 - b^2$ to compute $127 \times 113$ in your head.
Answer:
Factor $x^3 - 64$ using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.
x^3 - 64 = (x - p)(x^2 + qx + r)
A builder marks a right angle with a knotted rope. She measures $5$ m along one wall and $12$ m along the other. How long must the rope be between those two marks for the corner to be exactly square?
Answer:
You already collect like terms: $5x + 2x = 7x$. You already multiply a bracket by a number, and you already did long division with whole numbers — $\frac{372}{12}$, one place at a time, with a remainder left at the end. Polynomial arithmetic is those three habits applied to expressions instead of numbers, in the same order.
Degree: the highest power of $x$ that appears. $4x^3 - x + 1$ has degree $3$.
Leading coefficient: the number in front of the highest power.
Like terms: terms with exactly the same power of $x$. $3x^2$ and $-7x^2$ are like; $3x^2$ and $3x$ are not.
Dividend, divisor, quotient, remainder: in $\frac{372}{12} = 31$, $372$ is the dividend, $12$ the divisor, $31$ the quotient, $0$ the remainder. The same four words are used for polynomials.
Adding and subtracting is collecting like terms. Add $3x^2 - 2x + 5$ and $x^2 + 4x - 7$ power by power: $4x^2 + 2x - 2$. To subtract, change every sign inside the second bracket first.
Multiplying is the distributive law used until nothing is left: every term of one factor multiplies every term of the other, then collect. $(x + 3)(x^2 - x + 2) = x^3 - x^2 + 2x + 3x^2 - 3x + 6 = x^3 + 2x^2 - x + 6$. Two terms times three terms is six products, so count them before you collect.
Dividing is long division with powers of $x$ instead of powers of ten. To divide $x^2 + 5x + 6$ by $x + 2$: what times $x$ gives $x^2$? Just $x$. Subtract $x(x + 2) = x^2 + 2x$ and $3x + 6$ is left; what times $x$ gives $3x$? Just $3$; subtract $3(x + 2)$ and nothing is left, so the quotient is $x + 3$ and the remainder is $0$.
Adding, subtracting and multiplying polynomials always give polynomials. Dividing may not — which is why division is the one that leaves a remainder.
Another way: table
Multiplication as a grid. Write $x$ and $3$ down the side, $x^2$, $-x$ and $2$ along the top, and fill in the nine... six boxes: $x^3$, $-x^2$, $2x$ on the top row and $3x^2$, $-3x$, $6$ on the bottom. Adding the boxes down the diagonals collects the like terms for you, and no product can be forgotten because the grid has a box for each one.
Another way: story
Polynomials behave like whole numbers written in base $x$. $x^2 + 5x + 6$ is "$1$, $5$, $6$" the way $156$ is; adding, multiplying and dividing follow the same steps you learned for numbers, and the only thing that changes is that the columns never carry.
Adding the exponents when adding terms. $x^2 + x^2 = 2x^2$, not $x^4$. Exponents add when you multiply ($x^2 \times x^2 = x^4$), never when you add.
Dropping the minus sign after the first term. $(5x - 2) - (3x + 4)$ is $5x - 2 - 3x - 4 = 2x - 6$, not $5x - 2 - 3x + 4$. The minus applies to the whole bracket.
Forgetting the missing powers in a division. Dividing $x^3 - 7x + 6$ by $x - 1$, the $x^2$ term is missing — write it as $0x^2$ or the columns will not line up and every later term will be wrong.
Change every sign in the second bracket: $5x^3 - 2x + 1 - 2x^3 - x^2 + 4$.
The minus belongs to all three terms, not just the first.
Collect by power: $x^3$: $5 - 2 = 3$. $x^2$: $0 - 1 = -1$. $x$: $-2$. Numbers: $1 + 4 = 5$.
Go one power at a time so nothing is missed.
The answer is $3x^3 - x^2 - 2x + 5$.
$x^3 \div x = x^2$. Subtract $x^2(x - 1) = x^3 - x^2$, leaving $-x^2 + 3x - 5$.
One power at a time, highest first.
$-x^2 \div x = -x$. Subtract $-x(x - 1) = -x^2 + x$, leaving $2x - 5$.
$2x \div x = 2$. Subtract $2(x - 1) = 2x - 2$, leaving $-3$.
Nothing of degree $1$ or more is left, so we stop.
Quotient $x^2 - x + 2$, remainder $-3$.
Four products: $2x \times x = 2x^2$, $2x \times 4 = 8x$, $-1 \times x = -x$, $-1 \times 4 = -4$.
Collect the two $x$ terms: $8x - x = 7x$, so the answer is $2x^2 + 7x - 4$.
Simplify $(-6x^2 - 3x - 5) + (-2x^2 - 7x + 8)$.
Answer:
Expand $(x + 5)(x^2 + 6x + 8)$.
Answer:
Divide $x^2 - 14x + 48$ by $x - 8$.
Answer:
Simplify $(-5x^2 - 6x - 1) + (-2x^2 - 6x - 5)$.
Answer:
A rectangular garden is $3$ m longer than it is wide in one direction and $5$ m longer in the other: its sides are $x + 3$ and $x + 5$ metres. A square pond of side $x$ metres is dug in one corner. Write the area of grass left, in its simplest form.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Expand $(x - 9)^2$.
Answer:
Expand $(x + 3)(x^2 + 5x + 4)$.
Answer:
You can expand and factor with the standard identities and compute with polynomials. Expand $(x - 5)^2$ without writing the middle step, factor $x^3 - 64$, and divide $x^2 + 7x + 12$ by $x + 3$.
8. Your turn: compute $41^2$ without a calculator, step 2
19. Your turn: expand $(2x - 1)(x + 4)$, step 2