Up to now, $x$ has mostly shown up alone. In this section it gets squared, and that changes everything: equations can have two answers (or none), and graphs stop being straight lines and start curving. The ISEE tests the toolkit hard: multiplying things out, factoring them back, knowing when a fraction is zero versus broken, and reading a parabola. Every one of these has a classic trap, and we're going to name them all.
The practice is mixed into the lesson: each question sits right after the idea it needs, and the Mixed Practice at the bottom takes that hint away. Flip the switch on a question to type your answer instead: a number like 14, an expression like x^2 + x - 6, a product like (x + 3)(x - 2), or solutions like 2, 3 (type none when there aren't any). Questions whose choices are the answer, like "which graph?", stay multiple choice.
First, some names for the things we'll be multiplying. The big family comes first.
A polynomial is a sum of terms, each a number times variables raised to whole-number powers.
So $3x^2 - 5x + 2$ is a polynomial, and so is $x^2 + xy + y^2$. Polynomials with two terms show up constantly:
A binomial is a polynomial with two terms.
And the ones with three terms get their own name too:
A trinomial is a polynomial with three terms.
Counting terms is all it takes to tell them apart.
Which of these is a trinomial: $4x^2$, $x + 3$, $x^2 - 5x + 6$, or $x^3 - 2x^2 + x - 1$?
Count the terms, the pieces joined by $+$ and $-$. $4x^2$ has one, $x + 3$ has two (a binomial), $x^2 - 5x + 6$ has three, and $x^3 - 2x^2 + x - 1$ has four. The trinomial is $x^2 - 5x + 6$.
Now the main event: multiplying two binomials.
Multiplying two binomials is just the distributive property from Section 3, used twice: every term in the first factor multiplies every term in the second.
First, Outer, Inner, Last: four products, then combine like terms.
The cleanest way to see all four products is as the area of a box: one factor is the width, the other is the height.
The box is (x + q) wide and (x + s) tall. Red tiles add, blue tiles subtract.
The two x-tiles in the middle are where the trap lives. Here's how that plays out on a test question.
Which expression is equivalent to $(x - 2)(x + 3)$?
First: $x \cdot x = x^2$. Outer: $3x$. Inner: $-2x$. Last: $-2 \cdot 3 = -6$.
The middle terms combine: $3x - 2x = x$. So it's $x^2 + x - 6$. (Multiplying only the firsts and lasts, $x^2 - 6$, is the trap.)
Here's FOIL in slow motion, one product at a time.
Your turn. Watch the middle two products: they almost always combine.
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Bigger products work the same way: every term times every term.
Compare $(x - y)(x^2 + xy + y^2)$ with $x^3 - y^3$.
Distribute: $x(x^2 + xy + y^2)$ $= x^3 + x^2y + xy^2$ and $-y(x^2 + xy + y^2)$ $= -x^2y - xy^2 - y^3$.
The middle terms cancel in pairs, leaving exactly $x^3 - y^3$. They're equal, for every $x$ and $y$.
That cancelling is a pattern the ISEE loves. Try one.
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Three products come up so often they're worth memorizing.
The first one is the easiest to see. Build the square of side $a + b$ and count the pieces.
$(a + b)^2 = a^2 + b^2$ is called the Freshman's Dream. It's a dream because it isn't real.
Put the real square next to the dream and you can see exactly what goes missing.
Those two missing strips are the whole story behind a classic question.
If $m$ is a positive integer and $(x + 7)^2 = x^2 + mx + 49$, what is the value of $m$?
$(x + 7)^2$ $= x^2 + 2(7)x + 49$ $= x^2 + 14x + 49$, so $m = 14$. The middle term shows up twice (once from the outer product, once from the inner), which is why it's $2ab$.
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The third pattern is the friendliest: a sum times a difference of the same two terms. The Outer and Inner products cancel, and only the two squares are left.
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These patterns aren't just for letters.
These double as mental-math cheat codes. $49 \times 51$ $= (50 - 1)(50 + 1)$ $= 2500 - 1 = 2499$. And $51^2 = (50 + 1)^2$ $= 2500 + 100 + 1 = 2601$. Never square a sum by squaring the pieces: $51^2 \ne 2500 + 1$.
The trick is spotting which pattern a number problem is hiding. Sort these.
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The same patterns let you skip finding $x$ and $y$ altogether.
If $x^2 + y^2 = 30$ and $xy = 7$, what is the value of $(x + y)^2$?
Don't try to find $x$ and $y$. Expand instead: $(x + y)^2 = x^2 + 2xy + y^2$.
Regroup it as $(x^2 + y^2) + 2xy = 30 + 2(7) = 44$. (Forgetting the $2$ and answering $37$ is the trap.)
Whenever the givens are pieces of a special product, expand the product and drop the pieces in.
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The third pattern, read backward, has its own name.
A difference of squares is an expression of the form $a^2 - b^2$.
By the third special product, every difference of squares factors as $(a + b)(a - b)$. It has a picture too: a square with a smaller square cut out of the corner.
That picture settles a comparison the ISEE likes to ask.
Compare the shaded area above with $x^2 - xy - y^2$ (where $x > y > 0$).
The shaded area is the big square minus the corner: $x^2 - y^2$. The other expression takes away an extra $xy$, and $xy > 0$, so the shaded area is greater.
Now pick your own square and corner, and put the trick to work on a mental-math problem.
Cut the red strip and slide it around the side.
Factoring is multiplying in reverse.
Factoring an expression means rewriting it as a product.
For example, $x^2 + x - 6 = (x + 3)(x - 2)$: we just multiplied that one out, and factoring walks it back. Always start with the easy part:
Before anything else, pull out anything every term shares. Every term of $28n^4 + 16n^3 - 80n^2$ has a $4n^2$ in it, so it's $4n^2(7n^2 + 4n - 20)$. Skip this and the numbers get huge, and "factor completely" answers will look wrong to you.
Pulling out a number alone was Section 3's job. Here the common factor has an $x$ in it too.
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A difference of squares factors on sight: $x^2 - 25 = (x + 5)(x - 5)$. Just pull out any common factor first.
Factor completely: $2x^2 - 18$.
Both terms share a $2$: $2(x^2 - 9)$. Inside is a difference of squares, $x^2 - 3^2$.
So $2x^2 - 18 = 2(x + 3)(x - 3)$. (Stopping at $2(x^2 - 9)$ isn't "completely.")
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For trinomials, there's a method.
Factoring a trinomial is a factor-pair hunt, exactly like Same Area, New Shape in Section 0: list the pairs that multiply to a number, then pick the one that also adds up right.
With leading coefficient $1$, that hunt is the whole job. Here it is with tiles: the factors are the sides of a rectangle.
And here's the hunt itself: two numbers, one product, one sum.
Watch the signs: a negative product means one number of each sign.
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When the leading coefficient isn't $1$, a grid keeps the hunt organized.
To factor $ax^2 + bx + c$, make a 3-by-3 grid:
Here it is filling itself in for $7x^2 - 31x - 20$.
The grid works for leading coefficient $1$ too; it just fills itself in faster.
Factor $x^2 - 7x - 18$.
Top row: $1$, $-18$, $-18$. Two numbers that multiply to $-18$ and add to $-7$: $-9$ and $2$.
With $a = 1$, the rows are $1 \cdot (-9)$ and $1 \cdot 2$. Read along the X: $(x + 2)(x - 9)$.
Try it yourself, including one that refuses to cooperate.
Pick a quadratic, click the pair that works, then split a down column 1.
That last one was no accident.
TIC-TAC-TOE only works when the factors have integer coefficients. For $9x^2 - 7x - 56$, no two integers multiply to $-504$ and add to $-7$, so it can't be factored nicely. If you've checked every pair and nothing adds up, the answer isn't hiding: it doesn't exist. (Its equation still has solutions, though. The next part has a formula for them.)
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"Factor completely" means doing all of this, as many times as it takes: pull out a common factor, then split whatever is left, then check each new factor again. Peel one layer at a time.
Now we can solve equations with an $x^2$ in them.
A quadratic equation is an equation that can be written as $ax^2 + bx + c = 0$, with $a \neq 0$.
Factoring is how we solve them, thanks to one simple fact about zero:
If $A \cdot B = 0$, then $A = 0$ or $B = 0$.
Two numbers can only multiply to zero if one of them is zero. The answers we get this way have a name:
A root of an equation (or solution) is a value of $x$ that makes the equation true.
Put the two together: factor, then set each factor to zero.
Solve $x^2 - 5x + 6 = 0$.
Factor: $(x - 2)(x - 3) = 0$. Either $x - 2 = 0$ or $x - 3 = 0$, so $x = 2$ or $x = 3$. (Notice the signs flip: the factor $x - 2$ gives the root $+2$.)
Each factor is a switch: the product is $0$ exactly when one of them is.
A coefficient in front of $x$ just means one more step.
Solve $(3x + 2)(x - 5) = 0$.
$3x + 2 = 0$ gives $3x = -2$, so $x = -\frac{2}{3}$. And $x - 5 = 0$ gives $x = 5$.
So $x = -\frac{2}{3}$ or $x = 5$. (Answering $x = -2$ forgets to divide by the $3$.)
First the equations that come already factored...
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...then the ones you have to factor yourself.
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There's one catch, and it's the most common mistake on the whole page.
The zero-product property only works with zero. To solve $x^2 + x = 6$, you can't write $x(x + 1) = 6$ and set each piece equal to $6$. Move everything to one side first: $x^2 + x - 6 = 0$, so $(x + 3)(x - 2) = 0$ and $x = -3$ or $x = 2$.
Still tempted? Try both ways and check the answers.
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Some quadratics skip the factoring entirely:
$x^2 + 49 = 0$ has no real solution. The answer exists, just in a different universe. Multiverse movies were right.
In Section 10, that other universe gets a name: imaginary numbers. With $i = \sqrt{-1}$, the solutions of $x^2 + 49 = 0$ are $\pm 7i$.
Solve $x^2 = 49$. Then solve $x^2 + 49 = 0$.
$x^2 = 49$: both $7^2$ and $(-7)^2$ make $49$, so $x = \pm 7$. Don't lose the negative!
$x^2 + 49 = 0$ means $x^2 = -49$. But a real number squared is never negative, so there's no real solution.
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And when TIC-TAC-TOE fails, one formula never does:
gives the solutions of $ax^2 + bx + c = 0$. For $9x^2 - 7x - 56 = 0$, it gives $x = \dfrac{7 \pm \sqrt{2065}}{18}$: about $2.91$ or $-2.14$.
Quadratics also hide inside word problems. Name the unknown, build the equation, solve it, and then ask which answer actually makes sense.
A rectangle's length is $3$ more than its width, and its area is $40$. What is its width?
Call the width $w$, so the length is $w + 3$ and $w(w + 3) = 40$. Move everything to one side: $w^2 + 3w - 40 = 0$, which factors as $(w + 8)(w - 5) = 0$.
So $w = -8$ or $w = 5$. A width can't be negative, so the width is $5$ (and the length is $8$: check, $5 \cdot 8 = 40$).
Throwing out the answer that can't happen is part of the problem, not an afterthought.
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Fractions with $x$ on top and bottom add one more twist.
So to find where a fraction is zero, solve "top $= 0$" and then check each answer against the bottom.
For what value(s) of $x$ does $\dfrac{x^2 - 25}{(x + 2)(x - 3)}$ $= 0$?
The top is $0$ when $x^2 = 25$, so $x = \pm 5$. Neither one makes the bottom $0$, so both work: $x = -5$ or $x = 5$.
$x = -2$ and $x = 3$ make the bottom $0$: those make the fraction undefined, not zero. Picking them is the trap.
Sometimes the check throws out the only answer there is.
For what value of $x$ is $\dfrac{x + 3}{3 + x} = 0$?
The top and bottom are identical, so wherever the fraction exists it equals $1$. The only way to make the top $0$ is $x = -3$, but that makes the bottom $0$ too. No value works.
The sneakiest version hides the trap inside a factor that matches the top.
For what value(s) of $x$ does $\dfrac{x^2 - 9}{(x - 3)(x + 1)} = 0$?
The top is $0$ when $x^2 = 9$: $x = 3$ or $x = -3$.
But $x = 3$ also makes the bottom $0$, so the fraction is undefined there. Only $x = -3$ works. (Cancelling the $(x - 3)$ first doesn't make $x = 3$ allowed.)
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Feed the fractions some numbers and watch the top and the bottom separately.
Here's why that cancelled factor still matters: its graph has a hole.
For "undefined," ignore the top completely and hunt for every zero of the bottom.
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Graph a quadratic and you get a curve with a shape you already know.
The graph of $y = ax^2 + bx + c$ (with $a \neq 0$) is a U-shaped curve called a parabola.
Every basketball shot in every sports movie is a parabola. The slow-mo part? That's the vertex.
Every parabola turns around at exactly one point:
The vertex of a parabola is its turning point: the lowest point if it opens up, the highest point if it opens down.
And it's perfectly symmetric around that point:
The axis of symmetry is the vertical line through the vertex of a parabola.
Fold the parabola along that line and the two halves land on each other. Watch.
That graph's equation, $y = (x - 1)^2 - 4$, was written in a form that hands you the vertex.
If $a > 0$, it opens up (the vertex is a minimum). If $a < 0$, it opens down (the vertex is a maximum).
The sign inside the parentheses is where everyone slips.
What is the vertex of $y = 2(x + 4)^2 - 7$? Is its $y$-value a maximum or a minimum?
Write $x + 4$ as $x - (-4)$, so $h = -4$ and $k = -7$. The vertex is $(-4, -7)$.
$a = 2 > 0$, so it opens up and $-7$ is the minimum. (Reading the vertex as $(4, -7)$ is the sign trap.)
Build that one yourself, then flip it and push it off the axis.
Now go the other way: from the picture back to the equation.
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When the equation isn't in vertex form, the formula $x = -\frac{b}{2a}$ finds the vertex for you.
What is the minimum value of $y = x^2 + 8x + 10$?
The vertex is at $x = -\dfrac{b}{2a} = -\dfrac{8}{2} = -4$.
Plug it in: $y = (-4)^2 + 8(-4) + 10 = 16 - 32 + 10 = -6$. The minimum value is $-6$. (Plugging in $x = 0$ gives $10$, the $y$-intercept, not the minimum.)
Where does $-\frac{b}{2a}$ come from? Symmetry. Here's that same parabola.
The two points don't have to be at the $y$-intercept's height. Drag the line anywhere: the midpoint never moves.
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A line from Section 5 has the same slope everywhere. A parabola's steepness keeps changing: going down, flattening out at the vertex, then going up.
The roots of an equation are where its parabola crosses the $x$-axis, which is why a quadratic can have two solutions, one, or none:
Move the vertex yourself, and flip the parabola upside down: an upside-down one needs its vertex above the axis to cross it.
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The vertex is the highest or lowest point of the whole parabola. On a piece of it, that can change.
What is the maximum value of $y$ if $y = 2x^2 + 1$ for $-2 \le x \le 1$?
This parabola opens up, so its vertex, $(0, 1)$, is the minimum. The maximum is at whichever end of the interval is farthest from the vertex: $x = -2$, where $y = 2(4) + 1 = 9$.
Notice the vertex never got the maximum. That's the whole trick of these questions.
On an interval, always check the endpoints. For an upward parabola, the vertex is the minimum and the maximum sits at an endpoint. Answering with the vertex value ($1$) or the closer endpoint ($x = 1$ gives $3$) is exactly what the wrong choices are counting on.
Keep the parabola still and move the interval instead.
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That's quadratics: multiply out, factor back, solve, check the bottom of every fraction, and read the parabola.
Every question so far came right after its idea, which is a hint the real test never gives. Down here the hint is gone. First, the quantitative comparison: two columns, four fixed choices, and any idea from this section fair game. Watch for "cannot be determined": an equation like $x^2 = 36$ has two solutions, and they can land on opposite sides of a number.
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Next, the mixed set. Every question type on this page except the quantitative comparison goes into one hat, and each click draws a new one at random. Flip the switch to type your answer; the ones that need their choices stay multiple choice.
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Alright, if you've made it this far, it's time for something with a little heat. You'll build the quadratic yourself from a story, throw out the answer that can't happen, and take one more step. Sometimes TIC-TAC-TOE won't save you.
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