Factorising Flashcards

All 29 cards in this deck

How do you find the common factor to take out of an algebraic expression?

e.g. $15x^3 + 3x^2y$

Take the HCF of the number parts and the lowest power of each letter appearing in every term.

Here: $3x^2(5x + y)$

What are the steps to factorise by taking out a common factor?

e.g. $6x^2 + 15x$

1. HCF of the numbers: $3$
2. Lowest power of $x$ in every term: $x$, so factor is $3x$
3. Divide each term by $3x$: $3x(2x + 5)$

True or false? $x(9x + 6)$ is a full factorisation of $9x^2 + 6x$.

False. The bracket still has a common factor of $3$; the full factorisation is $3x(3x + 2)$.

How do you factorise an expression with a bracketed common factor?

e.g. $(x + y)^2 + 3(x + y)$

Treat the whole bracket as the common factor and take it out.

Here: $(x + y)(x + y + 3)$

True or false? When the common factor you take out is equal to one of the terms, that term becomes $1$ inside the bracket.

True.

e.g. $6x + 6 = 6(x + 1)$

What is factorising by grouping, and when is it used?

Used for four terms with no single common factor: pair the terms, take a common factor out of each pair, then take out the identical bracket.

What are the steps to factorise four terms by grouping?

e.g. $ax + bx - ay - by$

1. Pair the terms: $(ax + bx) + (-ay - by)$
2. Factorise each pair: $x(a + b) - y(a + b)$
3. Take out the common bracket: $(a + b)(x - y)$

True or false? When grouping, if the two brackets after factorising each pair are not identical, you cannot finish the factorisation that way.

True. You must re-pair the terms or take out a negative factor so the brackets match.

When grouping, what do you take out of a pair whose first term is negative?

e.g. $-ay - by$

A negative common factor, so the bracket comes out positive.

Here: $-y(a + b)$

To factorise $x^2 + bx + c$ as $(x + p)(x + q)$, what must $p$ and $q$ satisfy?

$p + q = b$ and $pq = c$.

e.g. for $x^2 + 5x + 6$ the numbers add to $5$ and multiply to $6$.

What are the steps to factorise a quadratic of the form $x^2 + bx + c$?

e.g. $x^2 + 5x + 6$

1. Find two numbers with product $c$ and sum $b$: product $6$, sum $5$ → $2$ and $3$
2. Put them in brackets with $x$: $(x + 2)(x + 3)$

True or false? To factorise $x^2 + 5x + 6$ the two numbers must add to $6$ and multiply to $5$.

False. It is the other way round: the sum must be $5$ and the product must be $6$.

In $x^2 + bx + c$ with $c$ negative, what are the signs of the two numbers in the brackets?

One positive and one negative.

e.g. $x^2 + 3x - 10 = (x + 5)(x - 2)$

In $x^2 + bx + c$ with $c$ positive and $b$ negative, what are the signs of the two numbers in the brackets?

Both negative.

e.g. $x^2 - 7x + 12 = (x - 3)(x - 4)$

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 $1 \times (-4) = -4$

What is the factorisation of $a^2 - b^2$?

$(a - b)(a + b)$

True or false? $x^2 + 9$ factorises as $(x + 3)(x + 3)$.

False. A sum of two squares does not factorise; only a difference of two squares does.

How do you factorise a difference of two squares when the squared term has a coefficient?

e.g. $9x^2 - 25$

Use the square root of each term in the brackets.

Here: $(3x - 5)(3x + 5)$

What are the steps to factorise fully a difference of two squares that has a common factor?

e.g. $50 - 2y^2$

1. Take out the common factor: $2(25 - y^2)$
2. Square root each term in the bracket: $5$ and $y$
3. Write as two brackets: $2(5 + y)(5 - y)$

Whatever the expression, what is the first thing to check before choosing a factorising method?

Whether every term has a common factor — take it out first.

Which method do you use for an expression of two terms that are both squares with a minus between them?

e.g. $p^2 - 49$

The difference of two squares: $(p - 7)(p + 7)$

Which method do you use for an expression of four terms with no factor common to all of them?

e.g. $ax + bx - ay - by$

Factorising by grouping: $(a + b)(x - y)$

Which method do you use for a three-term quadratic $ax^2 + bx + c$ where $a \neq 1$?

e.g. $6x^2 - 5x - 4$

Split the middle term using two numbers with product $ac$ and sum $b$, then group: $(2x + 1)(3x - 4)$

True or false? Once you have taken out a common factor, the factorisation is complete.

False. You must check whether the bracket factorises further.

e.g. $2m^2 - 2 = 2(m^2 - 1) = 2(m - 1)(m + 1)$

How can you check that a factorisation is correct?

Expand the brackets again and check you get the original expression.