OMPT drill — Algebra

Polynomials and factoring

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 six 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

Lesson

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, no exceptions, 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, and patterns only get fast through repetition, which is why this topic rewards drilling more than almost any other.

The good news is that the OMPT draws from a short pattern list. A common factor to pull out, a difference of squares, a three-term quadratic, and occasionally a perfect square: that covers nearly everything you will meet. The discipline that separates fast students from stuck ones is checking for a common factor first, every time, before trying anything cleverer. $3 x^{2} - 12$ looks unfactorable to someone hunting for two brackets, and becomes $3 (x - 2) (x + 2)$ the moment the 3 comes out.

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.