Factorising Harder Quadratics Flashcards

All 5 cards in this deck

When factorising $ax^2 + bx + c$ by splitting the middle term, what must the two numbers multiply and add to?

They multiply to $ac$ and add to $b$.

e.g. for $6x^2 - 5x - 4$: multiply to $-24$, add to $-5$.

What are the steps to factorise $ax^2 + bx + c$ by splitting the middle term?

e.g. $3k^2 + 11k - 4$

1. Two numbers with product $ac = -12$ and sum $11$: $12$ and $-1$
2. Split the middle term and group: $3k(k + 4) - 1(k + 4)$
3. Take out the common bracket: $(3k - 1)(k + 4)$

True or false? In a factorisation of $ax^2 + bx + c$, the first terms of the two brackets must multiply to give $ax^2$.

True.

e.g. for $6x^2$ the options are $2x \times 3x$ or $6x \times x$.

Before trying brackets for $ax^2 + bx + c$, what should you always check?

Whether all three terms have a common factor — take it out first.

e.g. $2x^2 + 10x + 12 = 2(x^2 + 5x + 6)$

How are the constants in the brackets of $ax^2 + bx + c$ related to $c$?

They multiply together to give $c$.

e.g. $6x^2 - 5x - 4 = (2x + 1)(3x - 4)$, and $1 \times (-4) = -4$