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A short class with Sarah and her students. The other ways are under More.
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Our equation is x² + 5x + 6 = 0. Which two numbers multiply to 6 and add to 5?
1 and 6 multiply to 6 but add to 7. 2 and 3 multiply to 6 and add to 5.
So the factorised form is (x + 2)(x + 3) = 0. How can you be sure it is right?
Expand it: x² + 3x + 2x + 6, which is x² + 5x + 6.
Good. Listing the factor pairs of c in order means you never miss the right pair.
And if c were bigger, like 24, I would list 1 and 24, 2 and 12, 3 and 8, 4 and 6.
We have (x + 2)(x + 3) = 0. If two numbers multiply to give zero, what must be true?
At least one of them has to be zero.
That is the null factor law. So x + 2 = 0 or x + 3 = 0.
That gives x = −2 or x = −3. The signs flip from the brackets.
Exactly. And why must the right-hand side be zero before we do this?
If (x + 2)(x + 3) equalled 2, the brackets could be lots of things, like 1 and 2. Only zero forces a factor to be zero.
Try x² − x − 12 = 0. What product and sum do we need?
Product −12 and sum −1. That is −4 and 3, so (x − 4)(x + 3) = 0 and x = 4 or x = −3.
Now 2x² − 8x = 0. Can I divide both sides by x?
No, we would lose x = 0. Take out 2x instead: 2x(x − 4) = 0, so x = 0 or x = 4.
Last one: 2x² + 7x + 3 = 0. Multiply a and c first.
a × c is 6, and 6 and 1 add to 7. Then 2x(x + 3) + 1(x + 3), which is (2x + 1)(x + 3).
So x = −½ or x = −3.
move every term to one side so the equation reads ax² + bx + c = 0 before factorising.
when a = 1, find two numbers that multiply to c and add to b.
if (x + p)(x + q) = 0, then x = −p or x = −q.
take out a common factor first, and spot a difference of two squares p² − q² = (p − q)(p + q).
find numbers that multiply to a × c and add to b, split the middle term and group.
the solutions are the x-intercepts of the parabola, and the axis of symmetry lies halfway between them.
Solve x² + 5x + 6 = 0: 2 × 3 = 6 and 2 + 3 = 5, so (x + 2)(x + 3) = 0 and x = −2 or x = −3.
Solve x² − x − 12 = 0: −4 × 3 = −12 and −4 + 3 = −1, so (x − 4)(x + 3) = 0 and x = 4 or x = −3.
Solve 2x² − 8x = 0: take out the common factor, 2x(x − 4) = 0, so x = 0 or x = 4.
Solve 2x² + 7x + 3 = 0: a × c = 6 = 6 × 1 and 6 + 1 = 7, so 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0 and x = −½ or x = −3.
Times to c and add to b, bracket up and set them free; each bracket zero, flip the sign, and both your roots will come out fine.
Two numbers go to a party called x² + 5x + 6. The host lets them in only if they multiply to 6 and add to 5. The pair 2 and 3 get in, and leave as the roots −2 and −3.
"Zero, Factor, Split, Solve" - make one side zero, factorise, split into two equations, solve each.
P.S.N. - Product, Sum, Null factor law: the three steps of solving by factoring.
Term
What do the two numbers need to do when factorising x² + bx + c?
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Term
State the null factor law.
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Solve (x + 2)(x + 3) = 0.
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How does x² − 49 factorise?
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Why should you not divide 2x² − 8x = 0 by x?
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An equation that can be written as ax² + bx + c = 0 with a ≠ 0.
To write an expression as a product of factors, such as x² + 5x + 6 = (x + 2)(x + 3).
If a product of factors is zero, then at least one factor is zero.
A value of x that makes the equation true; also called a solution.
An x-value where the graph meets the x-axis; the zeros of y = x² + 5x + 6 are −2 and −3.
An expression of the form p² − q², which factorises as (p − q)(p + q).
Rewriting bx as two terms whose coefficients multiply to a × c, so that the quadratic can be factorised by grouping.
Example 1 solves x² + 5x + 6 = 0 by factoring it into (x + 2)(x + 3) = 0, so x = −2 or x = −3. The lesson turns that one worked example into a reliable routine for solving any quadratic that factorises.
Factoring is the quickest way to solve a quadratic equation when it works. For x² + 5x + 6 = 0 we look for two numbers that multiply to 6 and add to 5, which are 2 and 3, so the equation becomes (x + 2)(x + 3) = 0. The null factor law then says one of the brackets must be zero, giving x = −2 or x = −3. This lesson works through that example in full, checks the answers, links them to the graph of y = x² + 5x + 6, and then applies the same routine to quadratics with negative terms, a common factor, a difference of two squares and a leading coefficient other than 1.
This lesson is the first of three worked examples on solving quadratic equations. Here the method is factoring; Example 2 uses completing the square and Example 3 uses the quadratic formula.
Example 1: Solve x² + 5x + 6 = 0.
Checking the answer. Expanding gives (x + 2)(x + 3) = x² + 3x + 2x + 6 = x² + 5x + 6, so the factorisation is right. Substituting also works: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0 and (−3)² + 5(−3) + 6 = 9 − 15 + 6 = 0.
What the answers mean on a graph. The solutions are the x-intercepts (zeros) of y = x² + 5x + 6, so the parabola crosses the x-axis at (−2, 0) and (−3, 0). The axis of symmetry lies halfway between them, at x = −2.5, and the vertex is at (−2.5, −0.25).
The same routine on other quadratics.
When factoring is the wrong tool. Some quadratics, such as x² + 4x + 1, have no pair of whole numbers that works because their solutions are irrational. For those, use completing the square (Example 2) or the quadratic formula (Example 3).