OMPT drill — Algebra

OMPT polynomials and factoring practice: lesson and 10 problems

Q: Is factoring on the OMPT-A?

Yes. The OMPT-A syllabus lists Algebra as a chapter of its own and Quadratic equations as another, and factoring runs through both. Expect roughly 4 to 6 of the 52 questions to ask you directly to expand, simplify or factor, and several quadratic questions where factoring is the quickest way to the solutions. Common factor, difference of squares and the two-number trinomial cover nearly all of it.

Q: What does "factor completely" mean on the OMPT?

Keep splitting until no bracket can be factored further over the integers. 3x^2 - 27 is not finished at 3(x^2 - 9); the answer is 3(x - 3)(x + 3). A sum of squares such as x^2 + 4 is where you stop, because it does not factor over the reals. Expand your final answer back in your head to confirm it matches the original.

Q: How do I factor a trinomial with a leading coefficient for the OMPT?

Use the ac method. For 6x^2 + 7x - 3, multiply 6 by -3 to get -18, then find two numbers with product -18 and sum 7: they are 9 and -2. Split the middle term into 9x - 2x, group the four terms in pairs, and the common bracket appears: (3x - 1)(2x + 3). It is slower than guessing when the numbers are tiny, but it never stalls.

Q: How many polynomial questions are on the OMPT-C?

The OMPT-C has only four chapters (Numbers, Algebra, Linear formulas and equations, Geometry) over 30 questions, so Algebra carries a large share. I would estimate 5 to 8 questions touch expanding or factoring directly, more than on any other variant in proportion. The questions are short: expand a product, factor a difference of squares or a simple trinomial, and give an exact answer.

Q: Why does dividing by x lose a solution on the OMPT?

Because x might be zero, and dividing by zero is not allowed. If you turn x^3 = 9x into x^2 = 9, you silently throw away the solution x = 0 and report two solutions instead of three. Move everything to one side, factor x out, and set each factor to zero: x(x - 3)(x + 3) = 0 gives all three. "How many solutions" questions are built around this slip.

Q: What is the difference between expanding and factoring on the OMPT?

They are inverse operations. Expanding removes brackets by multiplying every term by every term and collecting, and it always works mechanically. Factoring puts the brackets back, which requires recognising a pattern: a common factor, a difference of squares, a trinomial or a grouping. The OMPT asks both directions, so read the instruction word first and decide which way you are driving before you write.

Expanding brackets is mechanical; factoring is pattern recognition. Difference of squares and common-factor extraction cover most of what the OMPT actually asks. Work through all ten below before moving on; algebra slips compound into every other strand.

Tested inOMPT-A OMPT-B OMPT-C OMPT-D OMPT-E OMPT-F OMPT-G

OMPT polynomials and factoring: the lesson

What the OMPT tests in polynomials and factoring

Expanding and factoring are the same road driven in opposite directions. Expanding multiplies brackets out, every term in the first bracket times every term in the second, and it is purely mechanical. Factoring reverses it: you look at $2 x^{2} + 7 x - 15$ and reconstruct the brackets it came from. That direction is pattern recognition, which only gets fast through repetition. The Algebra chapter carries this on every variant, OMPT-A through OMPT-G, and the Quadratic equations chapter reuses it on all of them except OMPT-C, where it stays inside the Algebra questions.

The heavy sub-skills are expanding a product (including a squared bracket and a bracket times a trinomial), factoring by common factor, difference of squares and the two-number hunt for a trinomial, and solving a polynomial equation by factoring and setting each factor to zero. Lighter but real: trinomials with a leading coefficient other than 1, grouping, and reading a single coefficient off a product. The wording is compact: "Expand and simplify", "Factor completely", "Give the coefficient of $x^{2}$ ", "How many real solutions". The habit that separates fast students from stuck ones is checking for a common factor first, every time.

Notation and answer format the OMPT expects

Powers use ^ and products use *, with brackets typed as they read: (x-2)*(x+3). The star is never wrong.
The platform checks value, so $(x + 3) (x - 2)$ and $(x - 2) (x + 3)$ are both accepted. An expanded answer to a "factor" question is not.
"Factor completely" means keep going until no bracket splits further over the integers. $3 (x^{2} - 9)$ is not complete; $3 (x - 3) (x + 3)$ is.
A coefficient question wants a plain number, sign included: $- 17$ , not $- 17 x$ .
Solutions of an equation are listed with commas, in any order: x = -2, 2, 3. If a factor repeats, the solution is listed once.

Worked example 1: expand, then subtract a bracket

Expand and simplify $(3 x - 2)^{2} - (x + 1) (x - 4)$ .
The square is a pattern, not four products: $(3 x - 2)^{2} = 9 x^{2} - 12 x + 4$ . The middle term is twice the product of $3 x$ and $- 2$ .
The second product, in a fixed order so nothing goes missing: $x^{2} - 4 x + x - 4 = x^{2} - 3 x - 4$ .
Subtract the whole bracket, flipping every sign: $9 x^{2} - 12 x + 4 - x^{2} + 3 x + 4$ .
Collect: $8 x^{2} - 9 x + 8$ .
Check with $x = 2$ : the original is $4^{2} - 3 \cdot (- 2) = 16 + 6 = 22$ , and $8 \cdot 4 - 18 + 8 = 22$ . The slip to watch is subtracting only the first term of the second bracket.

Worked example 2: factor completely

Factor $2 x^{3} - 8 x^{2} - 24 x$ completely.
Common factor first, always: every term has $2 x$ in it, so $2 x (x^{2} - 4 x - 12)$ .
Inside is a trinomial with $x^{2}$ leading. I need two numbers with product $- 12$ and sum $- 4$ : $- 6$ and $2$ .
So $x^{2} - 4 x - 12 = (x - 6) (x + 2)$ , and the full answer is $2 x (x - 6) (x + 2)$ .
Check with $x = 1$ : original $2 - 8 - 24 = - 30$ ; factored $2 \cdot 1 \cdot (- 5) \cdot 3 = - 30$ . Stopping at $2 x (x^{2} - 4 x - 12)$ leaves the mark behind.

The full lesson works through trinomials with a leading coefficient, grouping, the zero-product rule, a parameter question and a word problem with a polynomial area. It ends with the edge cases the OMPT likes and the eight slips I mark most often.

How each OMPT variant tests polynomials and factoring

The chapter weights are the official ones from the omptest.org learning-assessment pages. The question counts are my estimates of how much of that chapter this one topic takes up. The 21-question papers (OMPT-D and OMPT-F) ask long multi-part questions, so a topic there usually shows up as a step inside a bigger problem.

OMPT-A
52 questions
Testedroughly 4 to 6 of 52 questions
Official chapter: Algebra, 15% of the paper.
Expand and simplify, factor completely, or give one coefficient of a product. Factoring is also the fast route through several Quadratic equations questions on this paper.
OMPT-B
64 questions
Testedroughly 4 to 6 of 64 questions
Official chapter: Algebra, 10% of the paper.
Same direct shapes as OMPT-A, plus factoring reused to simplify rational expressions and to prepare integrands, so the pattern list has to be automatic.
OMPT-C
30 questions
Testedroughly 5 to 8 of 30 questions
Official chapter: Algebra, 25% of the paper.
The Algebra chapter is a quarter of this short paper: expansions, common factors, difference of squares and simple trinomials as direct questions with exact answers.
OMPT-D
21 questions
Tested1 to 2 of the 21 long questions, usually as a step inside a bigger problem
Official chapter: Functions, equations and inequalities, 30% of the paper.
Rarely a bare factorisation. A cubic to reduce before solving, a rational function to simplify, or a limit of the form 0/0 that only gives way once the numerator is factored.
OMPT-E
40 questions
Testedroughly 3 to 4 of 40 questions
Official chapter: Algebra, 15% of the paper.
Direct expand and factor questions from the Algebra chapter, and factoring as the solving method inside the Quadratic equations chapter. Numeric or expression answers.
OMPT-F
21 questions
Tested1 to 2 of the 21 long questions, usually as a step inside a bigger problem
Official chapter: Algebra, 15% of the paper.
Factoring appears as one part of a function or differentiation question: find the zeros of a polynomial, simplify a derivative, or read a coefficient after an expansion.
OMPT-G
26 questions
Testedroughly 1 to 3 of 26 questions, mostly inside other chapters
Official chapter: Functions, graphs and equations, 45% of the paper.
No standalone Algebra chapter on this paper, so expanding and factoring are tested through quadratic equations, function questions and the algebra inside physics and chemistry items.

OMPT polynomials and factoring practice problems

10 original problems in the formats the OMPT uses: typed numeric answers, multiple choice, and spot-the-error. The last few are harder than the exam average on purpose. Solutions and the common mistake sit behind each problem.

Problem 1Multiple choice

Factor $z^{2} - 10 z + 21$ .

A $(z - 3) (z - 7)$
B $(z + 3) (z + 7)$
C $(z - 1) (z - 21)$
D $(z - 3) (z + 7)$

Show the worked solution

Answer: A — $(z - 3) (z - 7)$

I need two numbers that multiply to $+ 21$ and add to $- 10$ . Both must be negative: $- 3$ and $- 7$ work, since $(- 3) (- 7) = 21$ and $- 3 + (- 7) = - 10$ . So the factorisation is $(z - 3) (z - 7)$ . Expanding it back confirms the middle term $- 10 z$ .

Problem 2Numeric answer

Expand $(2 v - 3) (v + 4)$ and give the coefficient of the $v$ -term.

Show the worked solution

Answer: 5

Expanding term by term: $2 v \cdot v = 2 v^{2}$ , $2 v \cdot 4 = 8 v$ , $- 3 \cdot v = - 3 v$ , and $- 3 \cdot 4 = - 12$ . Collecting: $2 v^{2} + 5 v - 12$ . The coefficient of $v$ is $8 - 3 = 5$ .

Problem 3Numeric answer

How many distinct real solutions does $x^{3} - 4 x = 0$ have?

Show the worked solution

Answer: 3

I factor out $x$ first: $x (x^{2} - 4) = 0$ , and the difference of squares splits further: $x (x - 2) (x + 2) = 0$ . Three factors, three distinct roots: $x = 0$ , $x = 2$ , $x = - 2$ . So there are 3 solutions.

Problem 4Multiple choice

Which of these is a factor of $4 p^{2} - 25$ ?

A $2 p - 5$
B $4 p - 5$
C $2 p - 25$
D $p - 5$

Show the worked solution

Answer: A — $2 p - 5$

This is a difference of squares: $4 p^{2} = (2 p)^{2}$ and $25 = 5^{2}$ , so $4 p^{2} - 25 = (2 p - 5) (2 p + 5)$ . The factor listed is $2 p - 5$ . A quick sanity check: at $p = 2.5$ both the factor and the original expression equal zero.

Problem 5Spot the error

A student expands $(h - 6)^{2}$ . Step 1: $(h - 6)^{2} = h^{2} + 36$ . Step 2: so for $h = 1$ the value is $1 + 36 = 37$ . Which step contains the error?

Show the worked solution

Answer: Step 1

Step 1 squares each term separately and skips the cross term. The correct expansion is $(h - 6)^{2} = h^{2} - 12 h + 36$ . Step 2 then faithfully evaluates the wrong formula. With the right one, $h = 1$ gives $1 - 12 + 36 = 25$ , which matches $(1 - 6)^{2} = 25$ directly.

Problem 6Numeric answer

The expression $x^{2} + 9 x + 18$ factors as $(x + a) (x + b)$ with $a < b$ . What is $b - a$ ?

Show the worked solution

Answer: 3

I need $a \cdot b = 18$ and $a + b = 9$ . The pair $3$ and $6$ does both. With $a < b$ that means $a = 3$ , $b = 6$ , so $b - a = 3$ . Expanding $(x + 3) (x + 6)$ gives back $x^{2} + 9 x + 18$ , so the pair is right.

Problem 7Multiple choice

Factor $4 x^{2} - 4 x - 15$ completely.

A $(2 x - 5) (2 x + 3)$
B $(2 x + 5) (2 x - 3)$
C $(4 x - 5) (x + 3)$
D $(4 x + 3) (x - 5)$

Show the worked solution

Answer: A — $(2 x - 5) (2 x + 3)$

The leading coefficient is not 1, so I use the ac method: $a c = 4 \cdot (- 15) = - 60$ and I need two numbers with product $- 60$ and sum $- 4$ . Those are $- 10$ and $6$ . Split the middle term: $4 x^{2} - 10 x + 6 x - 15 = 2 x (2 x - 5) + 3 (2 x - 5) = (2 x - 5) (2 x + 3)$ . Check with $x = 1$ : original $4 - 4 - 15 = - 15$ ; factored $(- 3) (5) = - 15$ .

Problem 8Numeric answer

Expand $(3 x - 1) (x^{2} + 2 x - 5)$ and give the coefficient of the $x$ -term.

Show the worked solution

Answer: -17

Six products, in order. From $3 x$ : $3 x^{3} + 6 x^{2} - 15 x$ . From $- 1$ : $- x^{2} - 2 x + 5$ . Collect: $3 x^{3} + 5 x^{2} - 17 x + 5$ . The $x$ -term comes from $- 15 x - 2 x = - 17 x$ , so the coefficient is $- 17$ . Check with $x = 1$ : original $2 \cdot (- 2) = - 4$ ; expanded $3 + 5 - 17 + 5 = - 4$ .

Problem 9Spot the error

A student solves $(x - 2) (x + 5) = 8$ . Step 1: since the product is 8, either $x - 2 = 8$ or $x + 5 = 8$ . Step 2: so $x = 10$ or $x = 3$ . Step 3: the equation therefore has two solutions, 10 and 3. Which step contains the error?

Show the worked solution

Answer: Step 1

Step 1 applies the zero-product rule to a product that equals 8, not 0, and that rule only works for zero. The right route is to expand and bring everything to one side: $x^{2} + 3 x - 10 = 8$ , so $x^{2} + 3 x - 18 = 0$ , which factors as $(x + 6) (x - 3) = 0$ . The solutions are $x = - 6$ and $x = 3$ . Check: $(3 - 2) (3 + 5) = 8$ and $(- 6 - 2) (- 6 + 5) = (- 8) (- 1) = 8$ . The student got 3 by luck and 10 fails: $(8) (15) = 120$ .

Problem 10Numeric answer

For how many integer values of $k$ can $x^{2} + k x + 24$ be written as $(x + a) (x + b)$ with integers $a$ and $b$ ?

Show the worked solution

Answer: 8

Matching coefficients gives $ab = 24$ and $a + b = k$ . The positive factor pairs of 24 are $(1, 24), (2, 12), (3, 8), (4, 6)$ , with sums $25, 14, 11, 10$ . Both factors can also be negative, since a negative times a negative is still 24, and those pairs give sums $- 25, - 14, - 11, - 10$ . No mixed-sign pair works because the product would be negative. That is 8 distinct values of $k$ . Check one: $k = - 11$ gives $x^{2} - 11 x + 24 = (x - 3) (x - 8)$ .

OMPT polynomials and factoring: questions students ask

Is factoring on the OMPT-A?

Yes. The OMPT-A syllabus lists Algebra as a chapter of its own and Quadratic equations as another, and factoring runs through both. Expect roughly 4 to 6 of the 52 questions to ask you directly to expand, simplify or factor, and several quadratic questions where factoring is the quickest way to the solutions. Common factor, difference of squares and the two-number trinomial cover nearly all of it.

What does "factor completely" mean on the OMPT?

Keep splitting until no bracket can be factored further over the integers. $3 x^{2} - 27$ is not finished at $3 (x^{2} - 9)$ ; the answer is $3 (x - 3) (x + 3)$ . A sum of squares such as $x^{2} + 4$ is where you stop, because it does not factor over the reals. Expand your final answer back in your head to confirm it matches the original.

How do I factor a trinomial with a leading coefficient for the OMPT?

Use the ac method. For $6 x^{2} + 7 x - 3$ , multiply 6 by -3 to get -18, then find two numbers with product -18 and sum 7: they are 9 and -2. Split the middle term into 9x - 2x, group the four terms in pairs, and the common bracket appears: $(3 x - 1) (2 x + 3)$ . It is slower than guessing when the numbers are tiny, but it never stalls.

How many polynomial questions are on the OMPT-C?

The OMPT-C has only four chapters (Numbers, Algebra, Linear formulas and equations, Geometry) over 30 questions, so Algebra carries a large share. I would estimate 5 to 8 questions touch expanding or factoring directly, more than on any other variant in proportion. The questions are short: expand a product, factor a difference of squares or a simple trinomial, and give an exact answer.

Why does dividing by x lose a solution on the OMPT?

Because x might be zero, and dividing by zero is not allowed. If you turn $x^{3} = 9 x$ into $x^{2} = 9$ , you silently throw away the solution x = 0 and report two solutions instead of three. Move everything to one side, factor x out, and set each factor to zero: $x (x - 3) (x + 3) = 0$ gives all three. "How many solutions" questions are built around this slip.

What is the difference between expanding and factoring on the OMPT?

They are inverse operations. Expanding removes brackets by multiplying every term by every term and collecting, and it always works mechanically. Factoring puts the brackets back, which requires recognising a pattern: a common factor, a difference of squares, a trinomial or a grouping. The OMPT asks both directions, so read the instruction word first and decide which way you are driving before you write.

Test hubs that cover polynomials and factoring:OMPT-A OMPT-B OMPT-C OMPT-D OMPT-E OMPT-F OMPT-G All 28 topics

OMPT polynomials and factoring practice: lesson and 10 problems

OMPT polynomials and factoring: the lesson

What the OMPT tests in polynomials and factoring

Notation and answer format the OMPT expects

Worked example 1: expand, then subtract a bracket

Worked example 2: factor completely

How each OMPT variant tests polynomials and factoring

OMPT polynomials and factoring practice problems

OMPT polynomials and factoring: questions students ask

Related OMPT topics and test hubs