Combining like terms, factoring out the common factor, expanding expressions and factorising them, then the nine laws of exponents with worked examples on fractional powers and roots.

The first part of this lesson practises what the previous one introduced: combining like terms and taking a common factor outside the bracket. Note that the same expression can be factorised in several ways - each is correct, because expanding it brings back the same expression. The second part introduces powers: the names of the parts of the notation and the nine laws everything else follows from, including roots written as fractional powers.

Algebra - review exercises

Simplify each of the following expressions by combining like terms.

\text{a)}\quad 4xy + 3xz - 6zy - 5zx + yx = 5xy - 2zx - 6zy

\text{b)}\quad -2a + 4ab + a - 4ba = -a

\text{c)}\quad 3rst - 10str + 8ts - 5rt + 2st = -7rst + 10st - 5rt

\text{d)}\quad 2pq - 4pr + qr - 2rq + 3qp = 5pq - 4pr - qr

\text{e)}\quad 5lmn - 6ml + 7lm + 8mnl - 4ln = 13lmn + lm - 4ln

Simplify each of the following expressions by combining like terms and factoring out the common factor.

\begin{aligned}\text{a)}\quad 4xy + 3xz - 6zy - 5zx + yx &= 5xy - 2zx - 6zy = x(5y - 2z) - 6zy = \\&= 5xy - z(2x + 6y) = y(5x - 6z) - 2zx\end{aligned}

\begin{aligned}\text{b)}\quad 3rst - 10str + 8ts - 5rt + 2st &= -7rst + 10st - 5rt = t(-7rs + 10s - 5r) = \\&= st(-7r + 10) - 5rt = rt(-7s - 5) + 10st\end{aligned}

\begin{aligned}\text{c)}\quad 2pq - 4pr + qr - 2rq + 3qp &= 5pq - 4pr - qr = p(5q - 4r) - qr = \\&= 5pq - r(4p + q) = q(5p - r) - 4pr\end{aligned}

\begin{aligned}\text{d)}\quad 5lmn - 6ml + 7lm + 8mnl - 4ln &= 13lmn + lm - 4ln = l(13mn + m - 4n) = \\&= lm(13n + 1) - 4ln = ln(13m - 4) + lm\end{aligned}

Expand the following expressions, then factorise them where possible.

\begin{aligned}\text{a)}\quad 8x(y - z) + 2y(7x + z) &= 8xy - 8xz + 14xy + 2yz = 22xy - 8xz + 2yz = \\&= 2[x(11y - 4z) + yz]\end{aligned}

\begin{aligned}\text{b)}\quad (3a - b)(b - 3a) + b^{2} &= 3ab - 9a^{2} - b^{2} + 3ab + b^{2} = 6ab - 9a^{2} = \\&= 3a(2b - 3a)\end{aligned}

\begin{aligned}\text{c)}\quad -3\{w - 7[x - 8(3 - z)]\} &= -3\{w - 7[x - 24 + 8z]\} = \\&= -3\{w - 7x + 168 - 56z\} = \\&= -3w + 21x - 504 + 168z\end{aligned}

\begin{aligned}\text{d)}\quad \frac{2a - 3}{4b} + \frac{3a + 2}{6b} &= \frac{6a - 9 + 6a + 4}{12b} = \frac{12a - 5}{12b} = \\&= \frac{a}{b} - \frac{5}{12b} = \frac{1}{12b}(12a - 5)\end{aligned}

Powers

5a^{3}

5 is the coefficient, a is the base and 3 is the exponent.

The laws of exponents

\text{1)}\quad a^{m} \cdot a^{n} = a^{m+n}

\text{2)}\quad a^{m} : a^{n} = a^{m-n}

\text{3)}\quad (a^{m})^{n} = a^{m \cdot n}

\text{4)}\quad a^{0} = 1

\displaystyle \text{5)}\quad a^{-m} = \frac{1}{a^{m}}

\displaystyle \text{6)}\quad a^{\tfrac{1}{m}} = \sqrt[m]{a}

\text{7)}\quad a^{\tfrac{n}{m}} = \left(\sqrt[m]{a}\right)^{n} = \sqrt[m]{a^{n}}

\text{8)}\quad (ab)^{n} = a^{n} \cdot b^{n}

\displaystyle \text{9)}\quad \left(\frac{a}{b}\right)^{n} = \frac{a^{n}}{b^{n}}

Worked examples

\begin{aligned}\frac{6x^{-4} \cdot 2x^{3}}{8x^{-3}} &= \frac{6 \cdot 2x^{-4+3}}{8x^{-3}} = \frac{12x^{-1}}{8x^{-3}} = \frac{12}{8}x^{-1-(-3)} = \\&= \frac{3}{2}x^{-1+3} = \frac{3}{2}x^{2}\end{aligned}

\begin{aligned}E &= \left(5x^{2}y^{-\tfrac{3}{2}}z^{\tfrac{1}{4}}\right)^{2} \cdot \left(4x^{4}y^{2}z\right)^{-\tfrac{1}{2}} = \\&= 25x^{2 \cdot 2}y^{-\tfrac{3}{2} \cdot 2}z^{\tfrac{1}{4} \cdot 2} \cdot 4^{-\tfrac{1}{2}}x^{4 \cdot \left(-\tfrac{1}{2}\right)}y^{2 \cdot \left(-\tfrac{1}{2}\right)}z^{-\tfrac{1}{2}} = \\&= 25x^{4}y^{-3}z^{\tfrac{1}{2}} \cdot \frac{1}{2}x^{-2}y^{-1}z^{-\tfrac{1}{2}} = \\&= \frac{25}{2}x^{4+(-2)}y^{-3+(-1)}z^{\tfrac{1}{2}+\left(-\tfrac{1}{2}\right)} = \frac{25}{2}x^{2}y^{-4}z^{0} = \frac{25}{2}x^{2}y^{-4} \cdot 1\end{aligned}

E = \frac{25}{2}x^{2}y^{-4} = \frac{25x^{2}}{2y^{4}}

\begin{aligned}F &= \sqrt[3]{a^{6}b^{3}} : \sqrt{\frac{1}{9}a^{4}b^{6}} \cdot \left(4\sqrt{a^{6}b^{2}}\right)^{-\tfrac{1}{2}} = \\&= \left(a^{6}b^{3}\right)^{\tfrac{1}{3}} : \left(\frac{1}{9}a^{4}b^{6}\right)^{\tfrac{1}{2}} \cdot \left[4\left(a^{6}b^{2}\right)^{\tfrac{1}{2}}\right]^{-\tfrac{1}{2}} = \\&= a^{6 \cdot \tfrac{1}{3}}b^{3 \cdot \tfrac{1}{3}} : \frac{1}{3}a^{4 \cdot \tfrac{1}{2}}b^{6 \cdot \tfrac{1}{2}} \cdot \left(4a^{6 \cdot \tfrac{1}{2}}b^{2 \cdot \tfrac{1}{2}}\right)^{-\tfrac{1}{2}} = \\&= a^{2}b : \frac{1}{3}a^{2}b^{3} \cdot \left(4a^{3}b\right)^{-\tfrac{1}{2}} = \\&= 3b^{-2} \cdot \frac{1}{2}a^{3 \cdot \left(-\tfrac{1}{2}\right)}b^{-\tfrac{1}{2}} = \frac{3}{2}a^{-\tfrac{3}{2}}b^{-\tfrac{5}{2}} = \\&= \frac{3}{2\sqrt{a^{3}b^{5}}} = \frac{3}{2\sqrt{a^{2} \cdot a \cdot b^{4} \cdot b}} = \frac{3}{2ab^{2}\sqrt{ab}}\end{aligned}

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Frequently asked questions

Can an expression be factorised in only one way?

No. 5xy − 2zx − 6zy can be written as x(5y − 2z) − 6zy, as 5xy − z(2x + 6y), or as y(5x − 6z) − 2zx. Each one expands back to the original, so all of them are correct.

What are the parts of the notation 5a³ called?

5 is the coefficient, a is the base of the power and 3 is the exponent. The exponent says how many times the base appears as a factor, and the coefficient multiplies the whole power.

How do you write a root as a power?

The m-th root of a is a to the power of one over m, and the m-th root of a to the power n is a to the power n over m. That is why all the laws of exponents work for roots as well.

What is a number raised to the power of zero?

Always 1. You can see it in example E, where the exponents of z add up to zero and the whole z factor drops out of the result.

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Algebra: the laws of algebra, like terms and removing brackets

Logarithms: definition, laws and change of base