Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — expanding two brackets
Expand and simplify (2x+3)(x−5).
- 12x×x=2x2Every term in the first bracket multiplies every term in the second — four products in all.
- 22x×(−5)=−10xCarry the minus sign with the 5.
- 33×x=3x
- 43×(−5)=−15
- 52x2−10x+3x−15=2x2−7x−15Collect the two x terms: −10+3=−7.
Example 2 — factorising completely
Factorise 3x2−12 completely.
- 1Look for a common factor first: both terms divide by 3.Always the first move.
- 23x2−12=3(x2−4)Check by expanding: 3×x2=3x2 and 3×−4=−12 ✓
- 3x2−4 is a difference of two squares.x2−22, with a minus between them.
- 4=3(x−2)(x+2)Stopping at 3(x2−4) is not 'completely', and that is how the question is always worded.
Example 3 — changing the subject with a square
Make r the subject of A=πr2.
- 1r is multiplied by π and squared, so undo those in reverse order.Last operation applied is the first one undone.
- 2Divide both sides by π: πA=r2Undo the multiplication first.
- 3Take the square root of both sides.The opposite of squaring.
- 4r=πAThe whole fraction is under the root — dividing happened before the root is taken.