Review of Algebra
Theory
This Year 10 maths page revises the algebra from Years 7 to 9 that the rest of the course builds on: collecting like terms, multiplying and dividing terms, substitution with negative numbers, factorising with the highest common factor, and expanding brackets, including binomial products and a minus sign in front of a bracket.
A term is a number, a pronumeral, or numbers and pronumerals multiplied together, such as \(-4x^2y\). The number in front is the coefficient. The sign in front of a term belongs to it.
Like terms have exactly the same pronumeral part: \(3ab\) and \(-5ba\) are like terms, but \(x^2\) and \(x\) are not. Only like terms can be added or subtracted. To simplify, add or subtract their coefficients.
Any terms can be multiplied or divided. Multiply or divide the numbers, then the pronumerals: \(x^2 \times x = x^3\) and \(x^3 \div x = x^2\).
Substitution means replacing each pronumeral with its value. Put negative values in brackets.
Expanding removes brackets by multiplying every term inside by the term outside. Factorising is the reverse: take the highest common factor (HCF) outside the brackets. A binomial product is two brackets of two terms multiplied together.
The distributive law, for one bracket and for a binomial product:
Two special products:
How to simplify an algebraic expression
- Expand any brackets. Keep a product that follows a minus sign in brackets until it is expanded.
- Multiply or divide terms: numbers first, then each pronumeral. Divide the numbers by their HCF.
- Collect like terms, keeping each sign with its term.
- Order the answer: highest power first, then the number.
- Check by substituting a small value, such as \(x = 1\) or \(x = 2\), into the question and the answer.
How to substitute
- Replace each pronumeral with its value, in brackets if it is negative.
- Work out powers first, then multiply, then add and subtract.
(i) \(6y^2 - 4y + 2 - y^2 + 7y - 5\)
(ii) \(5pq - 3p + 2qp - p\)
Group the like terms. Each term keeps the sign in front of it.
(i)
| \(6y^2 - y^2\) | \(=\) | \(5y^2\) |
| \(-4y + 7y\) | \(=\) | \(3y\) |
| \(2 - 5\) | \(=\) | \(-3\) |
| \(\therefore\) | \(\) | \(\boldsymbol{5y^2 + 3y - 3}\) |
(ii) \(qp\) is the same as \(pq\), so \(5pq\) and \(2qp\) are like terms. \(p\) means \(1p\).
| \(5pq + 2pq\) | \(=\) | \(7pq\) |
| \(-3p - p\) | \(=\) | \(-4p\) |
| \(\therefore\) | \(\) | \(\boldsymbol{7pq - 4p}\) |
(i) \(-5x^2y \times 3xy\)
(ii) \(\dfrac{18p^3q}{12pq^2}\)
(i) Multiply the numbers, then the pronumerals. Add the powers of each pronumeral.
| \(-5x^2y \times 3xy\) | \(=\) | \(-5 \times 3 \times x^2 \times x \times y \times y\) |
| \(\) | \(=\) | \(\boldsymbol{-15x^3y^2}\) |
(ii) Divide the numbers by their HCF, \(6\). Cancel one \(p\) and one \(q\) from the top and the bottom.
| \(\dfrac{18p^3q}{12pq^2}\) | \(=\) | \(\dfrac{18}{12} \times \dfrac{p^3}{p} \times \dfrac{q}{q^2}\) |
| \(\) | \(=\) | \(\dfrac{3}{2} \times p^2 \times \dfrac{1}{q}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{3p^2}{2q}}\) |
Put each negative number in brackets. Work out each term on its own, then combine.
| \(3m^2\) | \(=\) | \(3 \times (-2)^2\) |
| \(\) | \(=\) | \(3 \times 4 = 12\) |
| \(-2mn\) | \(=\) | \(-2 \times (-2) \times 5 = 20\) |
| \(-n^2\) | \(=\) | \(-(5^2) = -25\) |
| \(\) | \(\) | \(3m^2 - 2mn - n^2\) |
| \(\) | \(=\) | \(12 + 20 - 25\) |
| \(\) | \(=\) | \(\boldsymbol{7}\) |
In \(3m^2\) only \(m\) is squared: \(3 \times 4\), not \((3 \times (-2))^2 = 36\).
Expand each part separately. Keep the second product in brackets, because the minus sign applies to all of it.
| \(4x(x + 2)\) | \(=\) | \(4x^2 + 8x\) |
| \((x + 3)(x - 5)\) | \(=\) | \(x^2 - 5x + 3x - 15\) |
| \(\) | \(=\) | \(x^2 - 2x - 15\) |
Subtract the whole second result:
| \(\) | \(\) | \((4x^2 + 8x) - (x^2 - 2x - 15)\) |
| \(\) | \(=\) | \(4x^2 + 8x - x^2 + 2x + 15\) |
| \(\) | \(=\) | \(\boldsymbol{3x^2 + 10x + 15}\) |
Check with \(x = 1\): the question gives \(4 \times 3 - 4 \times (-4) = 28\) and the answer gives \(3 + 10 + 15 = 28\).
Common pitfalls
Frequently asked questions
What are like terms in algebra?
Like terms have exactly the same pronumerals with the same powers, such as \(4xy\) and \(-2yx\). Only like terms can be added or subtracted.
How do you expand and simplify brackets?
Multiply each term inside the bracket by the term outside, watching the signs. Then collect the like terms. For two brackets, multiply each term in the first by each term in the second.
What does it mean to factorise fully?
Take out the highest common factor of every term, numbers and pronumerals. For \(18mn - 12n\) the HCF is \(6n\), so it factorises to \(6n(3m - 2)\).
How do you substitute a negative number?
Put the negative number in brackets, then work out powers first. For example, \((-2)^2 = 4\), so \(3m^2 = 12\) when \(m = -2\).
What is the difference between expanding and factorising?
They are opposites. Expanding removes brackets by multiplying. Factorising puts brackets back in by taking out a common factor.