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Read zeros, multiplicity and end behaviour off a factored polynomial, and use the remainder and factor theorems to find remainders, missing coefficients and the roots of a cubic.
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 a polynomial stops being an expression to compute with and becomes a curve you can describe. Each factor gives a zero, the power on that factor says whether the graph crosses the axis or only touches it, and the leading term alone decides what happens far out at either end. Then you meet the shortcut that makes higher-degree equations workable: the remainder on dividing $p(x)$ by $x - a$ is simply $p(a)$, so a factor is nothing more than a root, a missing coefficient becomes an equation, and a cubic with one root found by trial collapses into a quadratic you already know how to solve.
You already solve $(x - 2)(x + 5) = 0$ by saying that a product is zero only when one of its factors is zero, so $x = 2$ or $x = -5$. You already know that a parabola meets the $x$-axis where $y = 0$. A cubic or a quartic asks nothing new of you there: the same zero-product rule finds every crossing point. What is new is reading the shape off the factors — where the curve merely touches the axis, and what it does far out at either end.
Zero (or root): a value of $x$ at which $y = 0$; on the graph, an $x$-intercept.
Factor: one of the brackets the polynomial is a product of.
Multiplicity: how many times a factor is repeated. In $(x - 4)^2(x + 1)$ the zero $4$ has multiplicity $2$.
Degree: the highest power once expanded — and the number of zeros, counted with their multiplicities.
Leading coefficient: the number in front of that highest power.
End behaviour: what $y$ does as $x$ runs far out to the right, and far out to the left.
A polynomial written as a product of brackets tells you its graph almost without drawing.
Zeros. Each factor $(x - r)$ gives the zero $x = r$, because the product is zero when that bracket is. So $y = (x + 2)(x - 1)(x - 5)$ crosses at $-2$, $1$ and $5$, and a polynomial of degree $n$ has at most $n$ zeros.
Multiplicity. A repeated factor $(x - r)^k$ still gives one zero, but $k$ decides the shape there. Odd $k$: the factor changes sign as $x$ passes $r$, so the curve crosses. Even $k$: the sign is kept, so the curve touches and turns back. $y = (x - 1)^2(x + 3)$ touches at $1$ and crosses at $-3$.
End behaviour. Far out, the leading term is the whole story, because $x^n$ outgrows every lower power. For large positive $x$, $y$ takes the sign of the leading coefficient. On the left an even degree does the same as on the right, while an odd degree does the opposite: $y = x^3$ falls left and rises right, $y = x^4$ rises at both ends.
Another way: picture
The graph of $y = (x + 2)(x - 1)^2$: up from the bottom left, cutting the axis at $-2$, over a hump, down to kiss the axis at $1$ without going through, then away to the top right.
Another way: story
The factors are switches. Passing a zero of odd multiplicity flips $y$ across the axis; passing one of even multiplicity flips it twice, which is not at all.
Reading the zero straight off the bracket. $(x - 7)$ gives the zero $+7$, not $-7$. Set the bracket equal to zero and solve; the number crosses the equals sign with its sign changed.
"Every zero is a crossing." $y = (x - 3)^2$ touches the axis at $3$ and turns back, because a square is never negative. Odd multiplicity crosses; even multiplicity touches.
Judging the ends by the constant term, or by the degree alone. In $y = -2x^3 + 500x^2 - 9$ the term $500x^2$ leads for a while, but $-2x^3$ wins in the end, so the curve falls on the right. Degree is not enough either: the sign of the leading coefficient decides which way each end goes.
Set each bracket to zero: $x = -1$ and $x = 3$.
Two distinct zeros, each of multiplicity one.
Both multiplicities are odd, so the curve crosses the axis at both.
Expanded, the leading term is $x^2$: degree two, positive coefficient, so both ends rise. A parabola cutting the axis at $-1$ and $3$.
Even degree means the two ends agree.
The degree is $2 + 1 = 3$, and the leading coefficient is $-1$ because of the minus in front.
Count the powers in the brackets to get the degree.
Leading term $-x^3$. For large positive $x$ that is large and negative, so the curve falls without bound on the right.
At $x = 2$ the multiplicity is $2$, so the curve touches there; at $x = -4$ it is $1$, so it crosses.
Even touches, odd crosses.
Zeros at $1$ (multiplicity $1$, crosses) and $-2$ (multiplicity $2$, touches).
Degree $3$ with leading coefficient $+1$: falls on the left, rises on the right. So the curve rises from the bottom left, touches at $-2$, dips, then crosses upwards at $1$.
The graph of $y = (x - 1)(x - 2)(x - 8)$ meets the $x$-axis three times. Write the three zeros in increasing order.
x = z1, x = z2, x = z3
As $x$ grows large and positive, what does $y = 5x^3 + \ldots$ do? (The dots stand for terms of lower degree.)
The graph of $y = (x + 5)^2(x - 2)$ has a zero at $x = -5$. Does the curve touch the $x$-axis there and turn back, or cross it?
The graph of $y = (x + 6)(x - 3)(x - 8)$ meets the $x$-axis three times. Write the three zeros in increasing order.
x = z1, x = z2, x = z3
You have divided one polynomial by another and been left with a remainder, exactly as $\frac{47}{5}$ leaves $2$. You can substitute a number into an expression: if $p(x) = x^3 - 2x + 1$ then $p(3) = 27 - 6 + 1 = 22$. And you know that a factor is a divisor that leaves nothing over. This lesson connects those three: the remainder you would get by dividing is a number you can find by substituting instead.
Dividend, divisor, quotient, remainder: in $p(x) = (x - a)q(x) + R$, the polynomial $p$ is the dividend, $x - a$ the divisor, $q$ the quotient and $R$ the remainder.
$p(a)$: the number you get by putting $a$ in place of every $x$.
Remainder theorem: $R = p(a)$.
Factor theorem: $x - a$ is a factor of $p$ exactly when $p(a) = 0$.
Root: a solution of $p(x) = 0$; the same thing as a zero of $p$.
Divide a polynomial $p(x)$ by $x - a$. The remainder cannot contain $x$, because anything of degree $1$ or more would be divisible again, so it is a plain number $R$:
$$p(x) = (x - a)q(x) + R$$
That equation holds for every $x$, so it holds at $x = a$. There the first term dies, and
$$p(a) = 0 \cdot q(a) + R = R.$$
That is the remainder theorem: the remainder on dividing by $x - a$ is $p(a)$. Dividing $x^3 - 4x + 1$ by $x - 2$ leaves $8 - 8 + 1 = 1$, found without dividing anything.
The factor theorem is the case $R = 0$: $x - a$ is a factor of $p$ exactly when $p(a) = 0$ — exactly when $a$ is a root. This is the tool that cracks a cubic. Once you find one root by trial (try the divisors of the constant term), divide the cubic by its bracket, and what is left is a quadratic you can factor or solve with the formula. A cubic that looked unsolvable becomes a division and a quadratic.
The theorem also runs backwards. If a coefficient is unknown, the statement "$x - a$ is a factor" is an equation: write $p(a) = 0$ and solve it for the unknown.
Another way: table
Divide $p(x) = x^3 - 4x + 1$ by each of $x - 1$, $x - 2$, $x + 1$. Long division gives remainders $-2$, $1$, $4$; substitution gives $p(1) = -2$, $p(2) = 1$, $p(-1) = 4$. Same three numbers, one column obtained in a page and the other in a line.
Another way: story
Whole numbers behave the same way: the remainder of $n$ on division by $10$ is just the last digit, which you read off rather than compute. $p(a)$ is the polynomial's last digit in base $x - a$.
Substituting the wrong sign. For the divisor $x - 3$ you substitute $x = 3$; for $x + 3$ you substitute $x = -3$. Always ask what value makes the divisor zero, rather than copying the number you see.
"The remainder must be zero." It is zero only when the divisor is a factor. $p(x) = x^2 + 1$ divided by $x - 2$ leaves $p(2) = 5$, and that is a perfectly good answer.
Dividing when you were only asked for the remainder. Long division of a cubic takes a page and invites arithmetic slips; $p(a)$ takes one line. Divide only when you actually need the quotient — for instance when you are hunting for the remaining roots.
$x + 3$ is zero at $x = -3$, so the number to substitute is $-3$.
The sign flips; this is where most marks are lost.
$p(-3) = (-3)^3 + 2(-3)^2 - 5 = -27 + 18 - 5 = -14$.
Cube first, then square, then combine.
The remainder is $-14$, so $x + 3$ is not a factor.
Try the divisors of $6$. $p(1) = 1 - 6 + 11 - 6 = 0$, so $x - 1$ is a factor and $x = 1$ is a root.
One root found by substitution, not by algebra.
Divide: $x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)$.
The quotient is a quadratic, which is familiar ground.
$x^2 - 5x + 6 = (x - 2)(x - 3)$, so the roots are $1$, $2$ and $3$.
A factor means $p(2) = 0$: $8 + 2k - 10 = 0$.
$2k = 2$, so $k = 1$. Check: $8 + 2 - 10 = 0$.
Find the remainder when $x^3 + 2x + 4$ is divided by $x - 4$.
The remainder is answer.
$x - 2$ is a factor of $x^3 + kx + 18$. Find $k$.
k = answer.
One root of $x^3 + 6x^2 - 13x - 42 = 0$ is $x = -2$. Find the other two, smaller first.
x = u1 and x = u2
Find the remainder when $x^3 - 7x - 1$ is divided by $x - 3$.
The remainder is answer.
A quality engineer models the number of defects an assembly line produces at speed $x$ with a cubic $p(x)$. She never writes the cubic in her notebook. All she records is that dividing $p(x)$ by $x - 5$ leaves a remainder of $19$, and dividing $p(x)$ by $x - 4$ leaves a remainder of $11$. How many more defects does her model predict at speed $x = 5$ than at speed $x = 4$?
The model predicts answer more defects.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
As $x$ grows large and positive, what does $y = 3x^3 + \ldots$ do? (The dots stand for terms of lower degree.)
$x - 3$ is a factor of $x^3 + kx - 12$. Find $k$.
k = answer.
You can read a graph off a factored polynomial and use the remainder theorem. Say where $y = (x + 1)(x - 4)^2$ crosses and where it touches, and find the remainder when $x^3 - 2x + 5$ is divided by $x - 3$.
8. Your turn: sketch $y = (x - 1)(x + 2)^2$, step 2
19. Your turn: for what $k$ is $x - 2$ a factor of $x^3 + kx - 10$?, step 2