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Factor quadratics — common factors, trinomials, differences of squares and perfect squares — and solve them by factoring, square roots, completing the square and the discriminant.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you learn to run expansion backwards: to look at $x^2 + 5x + 6$ and find the two brackets it came from, taking out any common factor first and recognising the two patterns worth knowing by sight. You then use that to solve equations, because a product is zero only when one of its factors is — and you learn what to do when nothing factors neatly, including how the discriminant tells you in advance how many real solutions to expect.
You already expand $(x + 3)(x - 5)$ to $x^2 - 2x - 15$, and you already take a common factor out of $6x + 9$. Factoring a quadratic is that expansion run backwards: you are given the answer and asked for the two brackets, so the way to check is always to multiply back out.
Quadratic: a polynomial of degree $2$, such as $x^2 + 5x + 6$.
Factor: to write as a product. The reverse of expanding.
Greatest common factor (GCF): the largest thing dividing every term — a number, a power of $x$, or both.
Monic: a quadratic whose $x^2$ coefficient is $1$.
Perfect square trinomial: one that factors as a single bracket squared, like $x^2 - 10x + 25 = (x - 5)^2$.
Start every factorisation the same way: take out the greatest common factor. $6x^2 + 9x$ becomes $3x(2x + 3)$, and what is left is often much easier. For a monic quadratic $x^2 + bx + c$, look for two numbers whose product is $c$ and whose sum is $b$; they are the numbers in the brackets, because $(x + p)(x + q) = x^2 + (p + q)x + pq$. Two patterns are worth recognising rather than working out: a difference of squares $x^2 - k^2$ is $(x - k)(x + k)$ — you can tell it by the missing $x$ term — and a perfect square $x^2 \pm 2kx + k^2$ is $(x \pm k)^2$. When the leading coefficient is not $1$, as in $2x^2 + 7x + 3$, the same reasoning works on the expanded form $(ax + m)(x + n) = ax^2 + (an + m)x + mn$. Whatever route you take, expand your answer to check it.
Another way: diagram
A rectangle split into four cells for $(x + 2)(x + 3)$: $x^2$, $3x$, $2x$ and $6$, with the two $x$ cells shaded and labelled 'these add to $5x$' and the corner cell labelled 'this is $6$'.
Another way: story
Factoring is a small search, not a formula. Write down the factor pairs of the constant, then run down the list asking which pair adds to the middle coefficient. If the constant is negative, the pair has one of each sign.
"Find two numbers that add to the constant." The pair must multiply to the constant and add to the coefficient of $x$. Getting those two round the wrong way makes every factorisation fail.
"$6x^2 + 9x$ factors as $3(2x^2 + 3x)$." True but not complete: both terms also contain an $x$, so the answer is $3x(2x + 3)$.
"$x^2 + 25$ is a difference of squares." It is a sum of squares and does not factor over the real numbers. The pattern needs a minus sign.
Two numbers with product $6$ and sum $5$.
Product is the constant, sum is the middle coefficient.
The factor pairs of $6$ are $1 \times 6$ and $2 \times 3$; $2 + 3 = 5$.
Run down the list.
So $x^2 + 5x + 6 = (x + 2)(x + 3)$. Expanding gives back $x^2 + 5x + 6$.
The numbers share a factor of $3$, and both terms contain an $x$.
Always take the common factor first.
So the GCF is $3x$: $6x^2 \div 3x = 2x$ and $9x \div 3x = 3$.
Divide each term by it.
$6x^2 + 9x = 3x(2x + 3)$, and nothing is left to take out.
Two numbers with product $25$ and sum $-10$: both are $-5$.
So $x^2 - 10x + 25 = (x - 5)(x - 5) = (x - 5)^2$.
Factor $x^2 + 16x + 63$ as $(x + a)(x + b)$ with $a < b$. Fill in $a$ and $b$.
$a = $ a $,\quad b = $ b
Factor $30x^2 + 42x$ completely as $gx(ax + b)$. Fill in the three blanks.
$30x^2 + 42x = $ g $x($ a $x + $ b $)$
Factor $x^2 - 144$ as $(x - a)(x + b)$. Fill in $a$ and $b$.
$a = $ a $,\quad b = $ b
Factor $x^2 - 4x + 4$ as $(x - a)^2$. Fill in $a$.
$x^2 - 4x + 4 = (x - $ a $)^2$
You have just factored quadratics, you can take a square root, and you can solve a linear equation. Every method here reduces a quadratic to one of those: factoring turns it into two linear equations, square roots handle the case with no $x$ term, and the formula does the rest.
Root (or solution, or zero): a value of $x$ that makes the equation true.
Zero product property: if $AB = 0$ then $A = 0$ or $B = 0$. It is why factoring solves anything.
Discriminant: $b^2 - 4ac$. Positive means two real roots, zero means one, negative means none.
Completing the square: rewriting $x^2 + bx + c$ as $(x + \frac{b}{2})^2$ minus whatever that added.
Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Every method rests on one idea: if a product is zero, a factor is zero. So put the equation in the form (something) $= 0$, factor it, and set each bracket to zero: $x^2 - 5x + 6 = 0$ becomes $(x - 2)(x - 3) = 0$, giving $x = 2$ or $x = 3$. When there is no $x$ term, take square roots of both sides instead and keep both signs: $x^2 = 50$ gives $x = \pm\sqrt{50} = \pm 5\sqrt{2}$. When the quadratic does not factor neatly, complete the square — halve the coefficient of $x$, square it, and adjust — or use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, which is what completing the square gives in general. The discriminant $b^2 - 4ac$ under that root tells you in advance how many real solutions there will be: two if it is positive, one if it is zero, none if it is negative.
Another way: diagram
Three parabolas side by side: one cutting the $x$-axis twice labelled '$b^2 - 4ac > 0$', one touching it once labelled '$= 0$', and one floating above it labelled '$< 0$: no real roots'.
Another way: story
Pick the cheapest tool that works. No $x$ term: square roots. Factors easily: factor. Neither: the formula, which never fails and never surprises you.
"$(x - 2)(x - 3) = 6$, so $x - 2 = 6$ or $x - 3 = 6$." The zero product property is about zero only. Expand, move everything to one side, and factor again.
"$x^2 = 50$ gives $x = 5\sqrt{2}$." It gives $x = \pm 5\sqrt{2}$. Taking a square root of both sides always produces two cases.
"A negative discriminant means I made a mistake." It means the parabola never crosses the $x$-axis, so there is genuinely no real solution. That is an answer, not an error.
Two numbers with product $6$ and sum $-5$ are $-2$ and $-3$, so $(x - 2)(x - 3) = 0$.
Factor first.
A product is zero only if a factor is: $x - 2 = 0$ or $x - 3 = 0$.
The zero product property.
So $x = 2$ or $x = 3$. Both check in the original equation.
Half of $6$ is $3$, and $(x + 3)^2 = x^2 + 6x + 9$.
Halve the coefficient of $x$.
That is $7$ more than $x^2 + 6x + 2$, so the equation is $(x + 3)^2 - 7 = 0$.
Subtract what the square added.
So $(x + 3)^2 = 7$ and $x = -3 \pm \sqrt{7}$.
Square roots, both signs.
The discriminant is $b^2 - 4ac = 4 - 20 = -16$.
It is negative, so the equation has no real solutions.
Solve $x^2 + 13x + 42 = 0$. Fill in the solutions, smaller first.
$x = $ a $\ $ or $\ x = $ b
Solve $(x + 7)(x - 8) = 0$. Fill in the solutions, smaller first.
$x = $ a $\ $ or $\ x = $ b
Solve $x^2 = 162$ exactly, writing the answer as $x = \pm a\sqrt{2}$. Fill in $a$.
$x = \pm$ a $\sqrt{2}$
How many real solutions does $x^2 - 9x + 6 = 0$ have?
A rectangular garden covers $20$ m² and is $8$ m longer than it is wide. How wide is it?
answer
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Factor $2x^2 + 23x + 56$ as $(2x + m)(x + n)$. Fill in $m$ and $n$.
$m = $ m $,\quad n = $ n
Complete the square: $x^2 + 10x - 3 = (x + a)^2 - b$. Fill in $a$ and $b$.
$a = $ a $,\quad b = $ b
You can factor a quadratic and solve one by factoring, square roots or completing the square, and say how many real solutions it has. Explain why $(x - 2)(x - 3) = 6$ cannot be solved by setting each bracket to $6$.
8. Your turn: factor $x^2 - 10x + 25$, step 2
19. Your turn: how many real solutions has $x^2 + 2x + 5 = 0$?, step 2