Flashcards · Maths · Algebra
Factorising Expressions
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What is the first step in any factorising question?
Take out the highest common factor.
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Factorise \(6x + 15\).
\(3(2x + 5)\).
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Factorise fully \(6x^2y - 9xy^2\).
\(3xy(2x - 3y)\).
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How do you know an expression is fully factorised?
Nothing is left that divides into every term inside the brackets.
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What two numbers do you need to factorise \(x^2 + bx + c\)?
Two numbers that multiply to give \(c\) and add to give \(b\).
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Factorise \(x^2 + 7x + 12\).
\((x + 3)(x + 4)\).
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Factorise \(x^2 - 2x - 15\).
\((x - 5)(x + 3)\).
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How can you check a factorisation?
Expand the brackets and see whether you get the original expression back.
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What is the difference of two squares rule?
\(a^2 - b^2 = (a + b)(a - b)\).
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Factorise \(x^2 - 49\).
\((x + 7)(x - 7)\).
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Factorise \(4x^2 - 25\).
\((2x + 5)(2x - 5)\).
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Why does \(x^2 + 49\) not factorise?
It is a sum of squares. No pair of numbers multiplies to 49 and adds to 0.
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How can you work out \(51^2 - 49^2\) without a calculator?
\((51 + 49)(51 - 49) = 100 \times 2 = 200\).
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What is the factorised form of \(x^2 + 6x + 9\)?
\((x + 3)^2\), a perfect square.
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How do you factorise \(2x^2 + 7x + 3\) (Higher tier)?
Multiply \(2 \times 3 = 6\), split the middle term as \(6x + x\), and factorise in pairs: \((2x + 1)(x + 3)\).
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How do you simplify \(\dfrac{x^2 - 9}{x + 3}\) (Higher tier)?
Factorise the top to \((x + 3)(x - 3)\), cancel the common bracket and get \(x - 3\).
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Why can you not cancel the \(x\) in \(\dfrac{x + 6}{x}\)?
The \(x\) is only one term of the top, not a factor of the whole top, so the expression does not simplify to 6.