When factorising algebraic expressions with powers students often struggle to identify the highest common factor when it involves an algebraic term. For example, factorising 3h + 12 as 3(h + 4) is attempted correctly much more often than factorising 3h2 + 12h as 3h(h + 4). In this lesson students learn how to identify the highest common factor of expressions that include algebraic terms.
The starter activity helps students to understand factors are not restricted to numerical terms but could also include letters (or unknowns).
To help with this I have included some possible factor pairs of 36x2y. When I first taught this one student made an interesting point of including 72x2 and 1/2 y as a factor pair. Algebraically this works but we would not typically include fractions or decimals when finding the factor pairs of a number.
I ask students to work on mini-whiteboards to write at least 5 factor pairs of the term 40ab2c. For the purpose of this activity factor pairs involving fractions and decimals are encouraged as this comes up later in the lesson. A couple of the higher ability students include factor pairs such as 40abd and bcd-1 which I applaud.
When factorising terms involving powers students need to understand the basic rules of indices. As we work through the first couple of examples I write out each term as a product of the highest common factor. For example,
r3t + rt2
r3t = rt × r2
rt2 = rt × t
rt(r2 + t)
6w2y – 8wy2
6w2y = 2wy × 3w
– 8wy2 = 2wy × – 4y
2wy(3w – 4y)
8u3c2 – 20u2c
8u3c2 = 4u2c × 2uc
– 20u2c = 4u2c × -5
4u2c (2uc – 5)
I work through the first questions in this way and ask the class to attempt the next two on mini-whiteboards with similar working out. After this students work in their exercise book to match the equivalent expressions.
As students work independently through the questions on the third slide I challenge the most able to factorise expressions similar to those in the extension. I remember a similar question appearing in the final GCSE paper a couple of years back. The examiner’s report noted most students didn’t have a clue.
To factorise 16(f + d)2 + 8f + 8d most students only recognise 8 as a common factor. We discuss the need to factorise 8f + 8d and rewrite the expression as 16(f + d)2 + 8(f + d). It is now easier to see 8 and (f + d) are both common factors. 16(f + d)2 + 8f + 8d factorises to 8(f + d) (2f + 2d + 1).
In the plenary we investigate factorising the algebraic area of a rectangle to find possible perimeters. I ask everyone to include taking out the highest common factor for one of their solutions.
The plenary takes about 8 minutes with students working in pairs on mini-whiteboards.
For the less able students I emphasise the need to find the highest common denominator as they quite often only partially factorise by taking out the numerical factor.
For more able students I include problems involving negative powers such as 18x-2y + 27xy-2 = 9xy(2x-3 + 3y-3) This requires a greater understanding of the rules of indices.
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