VertexMath · MHF4U · Free worksheet

Polynomial Functions Review

Try each problem first. You don’t need to finish everything.

Core path ~55 min Full sheet ~1 h 45 min

Recommended: K1–K5, then one A and one T problem

K = Knowledge & Understanding · A = Application · T = Thinking · Boss = Challenge

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Quick reference

End behaviour: even degree, both ends go the same way; odd degree, opposite ways. A positive leading coefficient means the graph rises to the right.
Remainder and factor theorems: dividing $P(x)$ by $x-a$ leaves remainder $P(a)$ , and $x-a$ is a factor exactly when $P(a)=0$ . The factor $x+4$ goes with the zero $-4$ .
Order of a zero: at a zero of odd order the graph crosses the $x$ -axis; at a zero of even order it touches and turns back.
Cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$ and $a^3-b^3=(a-b)(a^2+ab+b^2)$ .
$(x-3)(x+4)>0$ is not solved factor by factor. Mark the zeros on a number line and test the sign of each interval.

Show your work on separate paper. Show your work in the dot-grid space.

Core Skills (K)

End Behaviour · Finite Differences · Factored Form · Division · Factoring

K1 End Behaviour & Symmetry

For each function, state the end behaviour, then determine whether the function is even, odd or neither.

a)
$f(x)=-3x^5$
b)
$g(x)=2x^4-5x^2+1$
c)
$h(x)=-2x^4+x$

Stuck? Peek at a hint

Replace $x$ with $-x$ and simplify: $f(-x)=f(x)$ means even, $f(-x)=-f(x)$ means odd.

K2 Finite Differences

The table shows values of a polynomial function $f$ .

$x$	$-1$	$0$	$1$	$2$	$3$	$4$
$f(x)$	$3$	$-1$	$1$	$-3$	$-25$	$-77$

a)
Determine the degree of $f$ using finite differences.
b)
Determine the leading coefficient of $f$ .
c)
State the end behaviour of $f$ and the maximum number of turning points its graph can have.

Stuck? Peek at a hint

Subtract consecutive $f(x)$ values, then repeat on each new row until one row is constant.

K3 Factored Form

Let $f(x)=-(x+2)^2(x-1)(2x-3)$ .

a)
State the zeros of $f$ and the order of each.
b)
Determine the degree, the leading coefficient and the $y$ -intercept.
c)
State the end behaviour and the zeros where the graph crosses the $x$ -axis.

Stuck? Peek at a hint

Set each factor equal to $0$ to read the zeros. For the leading coefficient, multiply the $x$ -coefficients of every factor, exponents included.

K4 Dividing Polynomials

a)
Use synthetic division to divide $x^4-3x^2+5x-7$ by $x+2$ . State the quotient and the remainder.
b)
Use long division to divide $6x^3+x^2-8x+7$ by $2x-1$ . Write the result as $P(x)=(2x-1)Q(x)+R$ .
c)
Without dividing, determine the remainder when $4x^3-2x^2+3x-1$ is divided by $2x+1$ .

Stuck? Peek at a hint

Before dividing, write a $0$ coefficient for every missing power of $x$ .

K5 Factoring Strategies

Factor fully.

a)
$27x^3+8$
b)
$2x^4-128x$
c)
$x^3-5x^2-9x+45$

Stuck? Peek at a hint

Take out any common factor first. For (c), group the first two terms and the last two terms.

Real World (A)

Polynomial Equations · Inequalities · Division in Context

A1 Open-Top Box

An open-top box is made from a 30 cm by 20 cm sheet of cardboard by cutting a square of side $x$ cm from each corner and folding up the sides.

a)
Write the volume $V(x)$ as a product of three factors and state its domain in interval notation.
b)
Find every cut size that gives a volume of $1000$ cm³. Round to the nearest tenth where needed.
c)
The equation $V(x)=1000$ has a third solution. Find it. Does it give a box?

Stuck? Peek at a hint

Set $V(x)=1000$ , bring every term to one side and divide by $4$ . Then test small integer values of $x$ with the factor theorem.

A2 Maple Syrup Profit

A maple syrup producer in Ontario models its yearly profit, in thousands of dollars, by $P(x)=-x^3+4x^2+7x-10$ , where $x$ is the number of bottles sold, in thousands. At high volumes, the extra labour and shipping cost more than the extra bottles earn.

a)
Factor $P(x)$ fully.
b)
Solve $P(x)>0$ . Give your answer in interval notation.
c)
Which part of your answer to (b) has no meaning for the producer? State the numbers of bottles that give a profit.

Stuck? Peek at a hint

Test the factors of $10$ in $P(x)$ to find a first zero. After factoring, mark the zeros on a number line.

A3 Storage Box

A rectangular storage box has volume $V(x)=2x^3+x^2-13x+6$ , in cubic centimetres, and height $(x+3)$ cm. Its length and width are also linear expressions in $x$ .

a)
Divide to find an expression for the area of the base.
b)
Factor the base area to find the length and width. Then find all three dimensions when $x=5$ .
c)
A designer wants the box to be $4$ cm tall. Is this possible? State every possible height.

Stuck? Peek at a hint

Divide $V(x)$ by $x+3$ ; synthetic division uses $-3$ .

Think It Through (T)

Error Analysis · Building Polynomials

T1 Spot the Flaw

A student factors $P(x)=x^3+3x^2-6x-8$ :
" $P(-1)=-1+3+6-8=0$ , so $(x-1)$ is a factor. Synthetic division by $1$ gives $x^2+4x-2$ , so $P(x)=(x-1)(x^2+4x-2)$ ."

a)
Evaluate $P(1)$ . What does the result tell you about $(x-1)$ ?
b)
Identify the two errors in the student's work.
c)
Factor $P(x)$ fully and state its zeros.

Stuck? Peek at a hint

For (a), substitute $x=1$ into $P(x)$ .

T2 Build It Backwards

Determine an equation for each polynomial function.

a)
The cubic function graphed below, which passes through the marked point $(2,-2.5)$ . Each grid square is 1 unit.
b)
A quartic whose graph touches the $x$ -axis at $x=-1$ and at $x=3$ , and passes through $(1,-32)$ .
c)
An odd function of least degree that has a zero at $x=2$ and passes through $(1,3)$ .

Stuck? Peek at a hint

Write one factor for each zero, with its order as the exponent, then use the given point to find $a$ .

Boss Level

Optional challenge

Two Remainders

When the polynomial $P(x)$ is divided by $x-1$ , the remainder is $3$ . When it is divided by $x+2$ , the remainder is $-3$ .

a)
When $P(x)$ is divided by $(x-1)(x+2)$ , the remainder has the form $ax+b$ . Find $a$ and $b$ .
b)
$P$ is a cubic with leading coefficient $1$ and $P(0)=7$ . Find $P(x)$ in standard form.
c)
Find the remainder when this $P(x)$ is divided by $x^2-1$ .

Stuck? Peek at a hint

Write $P(x)=(x-1)(x+2)Q(x)+ax+b$ , then substitute $x=1$ and $x=-2$ .

Answer key

Final answers only. Try first, then check.

Core Skills

a) As $x\to-\infty$ , $y\to\infty$ ; as $x\to\infty$ , $y\to-\infty$ ; odd
b) As $x\to\pm\infty$ , $y\to\infty$ ; even
c) As $x\to\pm\infty$ , $y\to-\infty$ ; neither

a) $3$ (third differences are constant: $-12$ )
b) $-2$
c) As $x\to-\infty$ , $y\to\infty$ ; as $x\to\infty$ , $y\to-\infty$ ; at most $2$ turning points

a) $-2$ (order 2), $1$ (order 1), $\frac{3}{2}$ (order 1)
b) Degree $4$ ; leading coefficient $-2$ ; $y$ -intercept $-12$
c) As $x\to\pm\infty$ , $y\to-\infty$ ; crosses at $x=1$ and $x=\frac{3}{2}$ (touches at $x=-2$ )

a) $Q(x)=x^3-2x^2+x+3$ ; $R=-13$
b) $6x^3+x^2-8x+7=(2x-1)(3x^2+2x-3)+4$
c) $-\frac{7}{2}$

a) $(3x+2)(9x^2-6x+4)$
b) $2x(x-4)(x^2+4x+16)$
c) $(x-5)(x-3)(x+3)$

Real World

a) $V(x)=x(30-2x)(20-2x)$ ; $D=(0,10)$
b) $x=5$ cm or $x=10-5\sqrt{2}\approx 2.9$ cm
c) $x=10+5\sqrt{2}\approx 17.1$ ; no: it is outside $(0,10)$ , and both $30-2x$ and $20-2x$ are negative there.

a) $P(x)=-(x+2)(x-1)(x-5)$
b) $(-\infty,-2)\cup(1,5)$
c) $(-\infty,-2)$ : negative sales are impossible. A profit needs more than $1000$ and fewer than $5000$ bottles.

a) $2x^2-5x+2$ cm² (remainder $0$ )
b) $(2x-1)$ cm and $(x-2)$ cm; at $x=5$ : $9$ cm by $3$ cm, height $8$ cm
c) No: a height of $4$ cm gives $x=1$ , so the width $x-2$ is $-1$ cm. All three dimensions are positive only when $x>2$ , so the height must be greater than $5$ cm.

Think It Through

a) $P(1)=-10$ : $(x-1)$ is not a factor; dividing by it leaves remainder $-10$ .
b) $P(-1)=0$ makes $(x+1)$ a factor, not $(x-1)$ ; and the division by $x-1$ left remainder $-10$ , which the student dropped.
c) $P(x)=(x+1)(x-2)(x+4)$ ; zeros $-4$ , $-1$ , $2$

a) $y=-\frac{1}{2}(x+3)(x-1)^2$
b) $y=-2(x+1)^2(x-3)^2$
c) $y=-x(x-2)(x+2)$ , that is $y=-x^3+4x$

Boss Level

Boss

a) $a=2$ , $b=1$ : the remainder is $2x+1$
b) $P(x)=x^3-2x^2-3x+7$
c) $-2x+5$

Free cheat sheet

Polynomial Functions Cheat Sheet

The whole unit on one page: every rule, a worked example for each idea, and the traps that cost marks.

Unit Summary

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Build your own

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Phone card

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Most testedFactor $P(x)=x^3-2x^2-5x+6$ fully and state its zeros. (card 5)

1End behaviour and symmetry

UseEven function: $f(-x)=f(x)$ , symmetric in the $y$ -axis. Odd: $f(-x)=-f(x)$ , symmetric about the origin.

Degree and leading coefficient $a$ (left end, right end):

even degree, $a>0$	up, up
even degree, $a<0$	down, down
odd degree, $a>0$	down, up
odd degree, $a<0$	up, down

TrapAn even degree does not make an even function: $x^4+x$ is neither.

2Differences and turning points

UseDegree $n$ : the $n$ th differences are constant and equal $a\cdot n!$ . At most $n-1$ turning points and $n$ zeros.

TrapConstant 3rd differences of $-12$ : $a=-12/3!=-2$ , not $-12$ .

3Zeros and their order

UseOdd order: the graph crosses. Even order: it touches and turns back. Leading coefficient: $a$ times the $x$ -coefficients. A family $y=a(\dots)$ : one point gives $a$ .

Example $y=-(x+2)^2(2x-1)$ : zeros $-2$ (touch) and $1/2$ (cross), leading coefficient $-2$ , $y$ -intercept $4$ .

TrapThe factor $(x+2)$ gives the zero $-2$ , not $2$ .

4Dividing polynomials

$P(x)=D(x)\,Q(x)+R$

UseWrite $0$ for each missing power. Synthetic division by $x-a$ uses $a$ : for $x+2$ use $-2$ ; for $2x+1$ use $-1/2$ , then divide the quotient by $2$ .

Trap $x^3-5x+2$ has no $x^2$ term: its coefficients are $1, 0, -5, 2$ .

5Remainder and factor theorems

Use $P(x)\div(x-a)$ leaves remainder $P(a)$ ; $x-a$ is a factor exactly when $P(a)=0$ . Test $\pm$ factors of the constant (over factors of $a$ ).

Example $P(x)=x^3-2x^2-5x+6$ : $P(1)=0$ , so $P=(x-1)(x^2-x-6)=(x-1)(x-3)(x+2)$ .

Trap $P(-1)=0$ means $x+1$ is a factor, not $x-1$ .

6Cubes and grouping

$a^3+b^3=(a+b)(a^2-ab+b^2)$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

UseCommon factor first. Four terms: group in pairs.

Trap $x^3+8=(x+2)(x^2-2x+4)$ : the middle sign is the opposite of the first.

7Solving equations

UseFactor completely and set each factor to $0$ . A quadratic factor with $D<0$ adds no real roots. $ax^4+bx^2+c$ : let $u=x^2$ . In context, keep only roots in the domain.

Example $x^4-13x^2+36=(x^2-4)(x^2-9)=0$ : $x=\pm2,\pm3$ .

TrapA box cut from a $20$ cm square with side $20-2x$ needs $0<x<10$ : reject roots outside.

8Inequalities

UseEverything on one side, factor, mark the zeros, test the sign of each interval. At an even-order zero the sign does not change.

Example $(x+1)(x-2)(x-5)>0$ : $(-1,2)\cup(5,\infty)$ .

TrapNot factor by factor: $(x-1)(x-5)>0$ is $x<1$ or $x>5$ , not " $x>1$ and $x>5$ ".

Which factoring?

Two cubes Cubes pattern

Four terms Grouping

No pattern Factor theorem

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