Pupil outcome: I can recognise when I can simplify before multiplying expressions by their respective terms. Key learning points: - Collecting together like terms simplifies the expression. - Simplifying may be possible before...

Pupil outcome: I can convert between metres and centimetres. Key learning points: - Using the known fact that 100 cm are equivalent to 1 m can help us convert between cm and m. - Partitioning and using the relationship between 1 m and...

Pupil outcome: I can find the n^th term formula for a quadratic sequence. Key learning points: - A quadratic sequence does not have a common difference. - The second differences between terms are equal. - The general n^th term formula is...

In this video, we work through a few examples of polynomial long division, and we relate it to long division of constants (regular numbers).

In this video, we do a quick few examples of factoring trinomials. To get a more conceptual understanding of why factoring by grouping works, please watch my other video: • Factor Polynomials by Grouping

In this video, we investigate how to factor by grouping when you are given four terms.

In this video, we factor polynomials with long division. Specifically, we look at examples where there is a "missing term," and we discover how to rewrite the polynomial so that it factors nicely.

In this video, we will factor polynomials that don't initially appear to be in the form of a quadratic. We will accomplish this by rewriting them in the form of a quadratic with substitution.

In this video, we factor special product binomials. Specifically, we factor difference of square, sum of cubes, and difference of cubes.

Find all six trig ratios given a point on the terminal side of theta. In this video we work this common type of problem. We use the Pythagorean Theorem to find "r" (hypotenuse), and then use the given "x" (adjacent) and "y" (opposite) to...

In this video, we use substitution to take equations and rewrite them in the form of a quadratic so that we can solve them.

In this video, we will work three problems using vertex form. 0:00 - rewriting standard form to vertex form. 3:25 - writing in vertex form when given vertex and another point on the parabola 5:54 - 2nd example of writing in vertex form...

In this video, we rewrite polynomials into factored form when given a zero.

In this video, we learn how to use explicit and recursive rules for sequences.

In this video, we solve rational equations by finding a common denominator.

How do you write the equation for an arithmetic sequence? This video answers this question. An arithmetic sequence is a sequence that adds or subtracts to find the next term (just like a linear function). We start by deriving the...

How do we add up the first "n" terms of a geometric sequence/series? In this video, we derive the equation to find the sum of a geometric series. Then, we use that rule to solve a few example problems where we add up the terms in a...

How do geometric sequences & series relate to repeating decimals? In this video we investigate how our knowledge of sum of infinite geometric series help us rewrite repeating decimals as fractions.

This is the first of my videos on factoring polynomials using the box method. Factoring polynomials is never easy, but I've seen several different strategies and I love the box method!

How do you factor polynomials with the box method when the leading coefficient isn't 1? This video explains how to factor this type of problem with a box.

This is the third of my videos on factoring polynomials using the box method. Factoring polynomials is never easy, but I've seen dozens of different strategies and the box method is the BEST

This video explains the connection between the box method and the distributive property when multiplying polynomials. The first example multiplies two binomials, while the second example multiples a trinomial and a binomial.

In this video, we simplify expressions by using the distributive property and combining like terms.

Find all six trig ratios given a point on the terminal side of theta. In this video we work this common type of problem. We use the Pythagorean Theorem to find "r" (hypotenuse), and then use the given "x" (adjacent) and "y" (opposite) to...