Look for this. Look for a repeated group or a common factor before testing possible zeros. An expression in x² can behave like a quadratic.
Group terms to reveal a shared binomial. In the remaining factor replace x² by u if all exponents are even. Factor back in x, keeping repeated factors.
Worked example
Factor P(x) = x⁵ − 2x⁴ − 5x³ + 10x² + 4x − 8 fully over the integers.
Group as x⁴(x − 2) − 5x²(x − 2) + 4(x − 2). Taking out x − 2 leaves x⁴ − 5x² + 4. With u = x² this is (u − 1)(u − 4). Each is a difference of squares after substitution. Hence P(x) = (x − 2)²(x + 2)(x − 1)(x + 1). The factor degrees add to 5.
Try it yourself
A design requires x + 1 to be a factor of Q(x) = x³ + ax² − 5x − 6. Find a and the remaining factors without graphing.
Check your reasoning
The factor theorem gives Q(−1) = −1 + a + 5 − 6 = 0, so a = 2. Divide x³ + 2x² − 5x − 6 by x + 1 to get x² + x − 6. Its factors are x + 3 and x − 2, since 3 and −2 have product −6 and sum 1.
Outcome RF12
Connect factors to the shape of a polynomial
Look for this. A graph that touches the axis and turns back has an even-multiplicity zero there. A crossing indicates odd multiplicity.
Add factor exponents to get the degree, then use its parity and the sign of the leading coefficient to orient both ends. Locate the zeros and check the y-intercept to pin down the vertical scale.
Worked example
Sketch the intercepts and end behaviour of P(x) = −(x + 2)²(x − 1)³. Explain what happens at each zero and give the range.
The degree is 5 with negative leading coefficient, so the left end rises and the right end falls. The graph touches at x = −2 and crosses with flattening at x = 1. It passes through (0, 4). Its continuous path connects both ends, so its range and domain are all real numbers.
Try it yourself
A polynomial graph touches the x-axis at −1 and 3, stays below the axis everywhere else, and passes through (0, −18). Find a least-degree formula and its absolute maximum.
Check your reasoning
Each touch requires at least a double factor, so start with a(x + 1)²(x − 3)². At x = 0 this is 9a = −18, giving a = −2. The formula is −2(x + 1)²(x − 3)². Its maximum value is 0, attained at both (−1, 0) and (3, 0).
Outcome RF13
Read a square-root graph from the graph underneath
Look for this. When a square root encloses a whole function, start by deciding where that function is non-negative.
Keep each allowed x-coordinate. Replace its old height y by √y. Heights of 0 and 1 stay fixed. Larger heights shrink. Parts below the x-axis disappear from the real graph.
Worked example
The inside function is f(x) = 9 − x². Sketch y = √f(x) using its domain, endpoints and highest point. Explain the shape.
The condition 9 − x² ≥ 0 gives [−3, 3]. The endpoints are (−3, 0) and (3, 0). The old peak (0, 9) becomes (0, 3). Squaring the new equation gives x² + y² = 9 with y ≥ 0, the upper semicircle. Its range is [0, 3].
Try it yourself
A sensor uses f(x) = x² − 4. Determine where √f(x) has real readings, where the reading is 1, and whether a sketch should join the left and right branches.
Check your reasoning
The radicand is non-negative when x ≤ −2 or x ≥ 2. For a reading of 1, solve x² − 4 = 1, giving x = ±√5. At x = ±2 the reading is 0. There are no real readings between −2 and 2, so draw two separate branches.
Outcome RF14
Distinguish a hole from a vertical asymptote
Look for this. A zero of the original denominator marks a forbidden input. Factoring reveals what the graph does nearby.
List excluded inputs before cancelling. If every copy of a denominator factor cancels, mark a hole at the reduced rule’s height. A factor left in the denominator gives a vertical asymptote.
Worked example
Sketch R(x) = (x² − 1)/(x² − 3x + 2). Mark each discontinuity, the horizontal asymptote and both intercepts.
Factor to (x − 1)(x + 1)/[(x − 1)(x − 2)], with x ≠ 1, 2. The reduced rule is (x + 1)/(x − 2) = 1 + 3/(x − 2). There is a hole at (1, −2), a vertical asymptote at x = 2 and a horizontal one at y = 1. The intercepts are (−1, 0) and (0, −½).
Try it yourself
A printer charges a $48 set-up fee and $2 per copy. Sketch the average cost per copy for positive whole copy counts. How many copies first bring the average below $5?
Check your reasoning
For n copies the average is (48 + 2n)/n = 2 + 48/n dollars. Plot separate points starting at (1, 50), approaching y = 2 from above. The inequality 48/n < 3 gives n > 16. The first count is 17, at about $4.82 per copy.
Worked notes and diagrams
1
The words, first
The idea: three families share one page, and each brings its own vocabulary that questions use without explaining.
Word
What it means
Polynomial function
A sum of terms in x with whole-number exponents: y = 2x³ − 5x² + 1.
Degree
The highest exponent. With the leading coefficient, it decides the end behaviour.
Leading coefficient
The number in front of the highest-degree term.
Zero (root)
A value of x where the graph meets the x-axis, so P(x) = 0.
Multiplicity
How many times a factor repeats. Odd multiplicity crosses the axis; even multiplicity touches and turns.
End behaviour
What the graph does far to the left and far to the right.
Remainder theorem
Dividing P(x) by x − a leaves a remainder of P(a).
Factor theorem
If P(a) = 0, then x − a is a factor.
Radical function
A function with the variable under a root, such as y = √(x − 2).
Invariant point
A point unchanged by a transformation — for square roots, where y = 0 or y = 1.
Rational function
A ratio of two polynomials, defined at inputs where the denominator is nonzero.
Vertical asymptote
A vertical line at a zero of the reduced denominator, where the rational function becomes unbounded.
Point of discontinuity
A break in a graph. For a rational function, a hole occurs when the denominator is nonzero at the excluded input after full cancellation.
2
Polynomial graphs
The idea: the degree and leading coefficient decide how the ends behave; the factors decide where the graph meets the x-axis and whether it crosses or bounces.
Odd multiplicity crosses the axis; even multiplicity touches it and turns back. Degree 3 with a positive leading coefficient: starts low, ends high.
Degree
Leading coefficient
Ends
odd
positive
down left, up right
odd
negative
up left, down right
even
positive
up on both ends
even
negative
down on both ends
Going the other way: a zero of −2 with multiplicity 2 and a zero of 3 gives y = (x + 2)²(x − 3). The sign inside each factor is the opposite of the zero.
3
The remainder and factor theorems
The idea: dividing P(x) by x − a leaves a remainder of P(a). If that remainder is 0, then x − a is a factor.
Remainder theorem: x² + 3x − 1 divided by x − 2 leaves P(2) = 4 + 6 − 1 = 9. No long division.
Factor theorem: for x³ − 7x + 6, P(1) = 1 − 7 + 6 = 0, so x − 1 is a factor.
Integral zero theorem: any integer zero must divide the constant term. For x³ − 2x² − 5x + 6, test ±1, ±2, ±3, ±6.
Finding an unknown coefficient
If x³ + kx² − 4 divided by x − 2 leaves 8, then P(2) = 8: 8 + 4k − 4 = 8, so k = 1.
Factoring a cubic
Test the possible integer zeros until one gives 0 — say x = 1 for x³ − 2x² − 5x + 6.
Divide by x − 1 (synthetic division is fastest) to get x² − x − 6.
Factor what is left: (x − 3)(x + 2). So P(x) = (x − 1)(x − 3)(x + 2).
Watch out: dividing by x + 2 means a = −2, so the remainder is P(−2), not P(2).
4
Radical functions
The idea: y = √x starts at (0, 0) and only goes right and up, because you cannot square-root a negative. Every radical question is about that starting point and that restriction.
Transformations work exactly as in the transformations unit. For y = √(x − 2) + 3, the starting point moves from (0, 0) to (2, 3), so the domain is x ≥ 2 and the range is y ≥ 3.
From y = f(x) to y = √f(x)
Wherever f(x) is negative, √f(x) does not exist.
Points where y = 0 or y = 1 stay put — those are the invariant points, since √0 = 0 and √1 = 1.
Between 0 and 1 the graph rises (√0.25 = 0.5); above 1 it drops (√9 = 3).
Solving radical equations
Isolate the root, square both sides, solve — then check every answer. For √(x + 5) = x − 1, squaring gives x² − 3x − 4 = 0, so x = 4 or −1. But x = −1 makes the left side √4 = 2 and the right side −1 − 1 = −2, so the two sides disagree and −1 is extraneous. Only x = 4 works.
Watch out: squaring creates extraneous roots. A square root is never negative, so any answer that makes the right side negative has to go.
5
Rational functions
The idea: where the denominator is zero, the graph breaks. Factor and reduce completely. A finite reduced value gives a hole. A surviving denominator zero gives a vertical asymptote.
The graph shoots off near x = 3 and flattens toward y = 0 far from the origin, but never touches either line.
Asymptote or hole?
For y = (x − 2) / ((x − 2)(x + 1)):
x − 2 cancels → a hole (point of discontinuity) at x = 2.
x + 1 does not cancel → a vertical asymptote at x = −1.
Horizontal asymptotes
A horizontal asymptote describes end behaviour. For example, y = x/(x² + 1) approaches y = 0 at both ends and crosses that line at (0, 0).
Degree on top is lower than the bottom → y = 0. Example: y = 1/x.
Same degree on top and bottom → the ratio of leading coefficients. For y = (2x + 1)/(x − 3), it is y = 2.
Watch out: calling every zero of the denominator an asymptote. In (x − 2)/(x − 2)², one copy cancels but 1/(x − 2) remains. The vertical asymptote is x = 2.
Practice by topic
Work on paper, choose an answer, then compare your method with the explanation.
Polynomial functions · 5 questions
Practice 1
What is the degree of y = 3x⁴ − 2x + 7?
1
7
4
3
Answer and explanation
4
The degree is the highest power of x, which is 4. The 3 is the leading coefficient and the 7 is the constant term.
Practice 2
What are the zeros of y = (x − 2)(x + 5)?
−2 and 5
2 and 5
−10
2 and −5
Answer and explanation
2 and −5
Set each factor to zero: x − 2 = 0 gives 2 and x + 5 = 0 gives −5. The signs flip from what you see in the brackets.
Practice 3
Which describes the end behaviour of y = x³?
down to the left, up to the right
up on both ends
up to the left, down to the right
down on both ends
Answer and explanation
down to the left, up to the right
Odd degree means the ends go opposite ways; a positive leading coefficient means it rises to the right. So it starts low and ends high.
Practice 4
At x = 1, the graph of y = (x − 1)²(x + 3):
has a hole
crosses the x-axis
has a vertical asymptote
touches the x-axis and turns around
Answer and explanation
touches the x-axis and turns around
The factor x − 1 is squared, so x = 1 is a zero of multiplicity 2. Even multiplicity touches and bounces back; odd multiplicity crosses.
Practice 5
Which polynomial of least degree has a zero of −2 with multiplicity 2 and a zero of 3?
y = (x − 2)²(x + 3)
y = (x + 2)(x − 3)
y = (x + 2)(x − 3)²
y = (x + 2)²(x − 3)
Answer and explanation
y = (x + 2)²(x − 3)
A zero of −2 comes from the factor x + 2, and multiplicity 2 squares it. A zero of 3 comes from x − 3. That gives degree 3.
Remainder and factor theorems · 5 questions
Practice 1
What is the remainder when P(x) = x² + 3x − 1 is divided by x − 2?
9
1
−11
3
Answer and explanation
9
The remainder theorem says the remainder is P(2): 4 + 6 − 1 = 9. No long division needed.
Practice 2
If P(4) = 0, which statement must be true?
4 is the remainder when P(x) is divided by x
x + 4 is a factor of P(x)
x − 4 is a factor of P(x)
P(x) has degree 4
Answer and explanation
x − 4 is a factor of P(x)
The factor theorem: if P(a) = 0, then x − a is a factor. Here a = 4, so x − 4.
Practice 3
Which is a factor of P(x) = x³ − 7x + 6?
x + 1
x − 3
x + 2
x − 1
Answer and explanation
x − 1
Test each: P(1) = 1 − 7 + 6 = 0, so x − 1 is a factor. P(−1), P(3) and P(−2) all give 12, not 0.
Practice 4
When x³ + kx² − 4 is divided by x − 2, the remainder is 8. What is k?
−1
3
1
2
Answer and explanation
1
The remainder is P(2): 8 + 4k − 4 = 8. So 4 + 4k = 8, and k = 1.
Practice 5
By the integral zero theorem, which are the possible integer zeros of P(x) = x³ − 2x² − 5x + 6?
1, 2, 3 and 6 only
±1, ±2, ±5
±1, ±5
±1, ±2, ±3, ±6
Answer and explanation
±1, ±2, ±3, ±6
Integer zeros must divide the constant term, 6, and can be positive or negative: ±1, ±2, ±3, ±6. (1, −2 and 3 turn out to work.)
Radical functions · 4 questions
Practice 1
What is the domain of y = √x?
x ≥ 0
x ≤ 0
all real numbers
x > 0
Answer and explanation
x ≥ 0
You cannot take the square root of a negative number, but √0 = 0 is fine, so x ≥ 0.
Practice 2
What is the range of y = √(x − 2) + 3?
y ≥ 2
y ≥ 0
y ≥ 3
y ≥ −3
Answer and explanation
y ≥ 3
√ is never negative, so the smallest value of the root is 0. Add 3 and the smallest y is 3. The −2 inside affects the domain, not the range.
Practice 3
When graphing y = √f(x) from y = f(x), the invariant points occur where:
x = 0 and x = 1
the graph crosses the y-axis
y = 0 and y = 1
y = 1 only
Answer and explanation
y = 0 and y = 1
√0 = 0 and √1 = 1, so points with those y-values stay exactly where they are. Everywhere else the y-value changes.
Practice 4
Solve √(x + 5) = x − 1.
no solution
x = 4 and x = −1
x = −1 only
x = 4 only
Answer and explanation
x = 4 only
Squaring gives x + 5 = x² − 2x + 1, so x² − 3x − 4 = 0 and x = 4 or −1. Check: x = −1 gives √4 = 2 but −1 − 1 = −2, so it is extraneous.
Rational functions · 4 questions
Practice 1
What is the vertical asymptote of y = 1/(x − 3)?
y = 0
x = 3
y = 3
x = −3
Answer and explanation
x = 3
The function is undefined where the denominator is zero: x − 3 = 0, so x = 3. y = 0 is the horizontal asymptote.
Practice 2
What is the horizontal asymptote of y = 1/x?
y = 0
y = 1
x = 0
there is none
Answer and explanation
y = 0
As x gets very large or very negative, 1/x gets closer and closer to 0 without reaching it. x = 0 is the vertical asymptote.
Practice 3
What does the graph of y = (x − 2) / ((x − 2)(x + 1)) have?
a hole at x = −1 and a vertical asymptote at x = 2
a hole at x = 2 and a vertical asymptote at x = −1
vertical asymptotes at x = 2 and x = −1
no asymptotes and no holes
Answer and explanation
a hole at x = 2 and a vertical asymptote at x = −1
x − 2 cancels from top and bottom, which leaves a hole (point of discontinuity) at x = 2. x + 1 does not cancel, so x = −1 is a vertical asymptote.
Practice 4
What is the horizontal asymptote of y = (2x + 1)/(x − 3)?
y = −1/3
y = 2
x = 3
y = 0
Answer and explanation
y = 2
The top and bottom have the same degree, so the asymptote is the ratio of leading coefficients: 2/1 = 2. x = 3 is the vertical asymptote.
maddyhelps.