Factored Form of a Quadratic

Vertex form shows you the vertex. Factored form, $y = a(x - r)(x - s)$, shows you something different: where the parabola crosses the $x$-axis. Once you know the zeros, symmetry hands you the axis and the vertex almost for free. This is where all your factoring practice pays off.

Key ideas

Factored form shows the zeros

A quadratic relation in factored form looks like

$$y = a(x - r)(x - s)$$

The zeros are where $y = 0$. A product is $0$ only when one of its factors is $0$, so $y = 0$ when $x - r = 0$ or $x - s = 0$. The zeros are $r$ and $s$.

Watch the signs. Each zero is the number that makes its bracket equal $0$:

• $y = (x - 5)(x + 3)$ has zeros $5$ and $-3$.
• $y = 2x(x - 4)$ has zeros $0$ and $4$ (the factor $x$ is $0$ when $x = 0$).
• $y = (2x - 3)(x + 1)$ has zeros $\tfrac{3}{2}$ and $-1$ (solve $2x - 3 = 0$).

Factors and x-intercepts go together

This works both ways:

• every factor $(x - r)$ gives an $x$-intercept at $r$
• every $x$-intercept $r$ gives a factor $(x - r)$

So the graph and the factors are two views of the same information. If a quadratic has no $x$-intercepts, it can’t be written as $a(x - r)(x - s)$ with real numbers $r$ and $s$.

Axis of symmetry and vertex

The axis of symmetry is halfway between the zeros:

$$x = \frac{r + s}{2}$$

To find the vertex, substitute that $x$-value into the equation to get the $y$-coordinate. That $y$-value is the minimum (if $a \gt 0$) or maximum (if $a \lt 0$).

The y-intercept and the direction of opening

• The $y$-intercept comes from setting $x = 0$.
• The parabola opens up if $a \gt 0$ and down if $a \lt 0$, just like in every other form.

Sketching from factored form

1. Plot the zeros $r$ and $s$ on the $x$-axis.
2. Find the axis of symmetry halfway between them.
3. Substitute to find the vertex, and plot it.
4. Find and plot the $y$-intercept (and its mirror image, if it helps).
5. Draw a smooth parabola through the points.

The sketch of $y = 0.5(x + 2)(x - 4)$ from Example 2.

From standard form to factored form

To use factored form, you often have to get there first by factoring $y = ax^2 + bx + c$:

1. Take out any common factor first.
2. Factor the trinomial that’s left, using simple trinomials, complex trinomials, or special cases such as a difference of squares.

Finding the equation from the zeros and a point

If you know the zeros $r$ and $s$ and one other point:

1. Write $y = a(x - r)(x - s)$ with the zeros filled in.
2. Substitute the other point and solve for $a$.

Worked examples

Example 1: Reading the features

For $y = 2(x - 1)(x + 3)$, find the zeros, the axis of symmetry, the vertex, the $y$-intercept, and the minimum or maximum value.

Solution.

Zeros. $x - 1 = 0$ gives $x = 1$, and $x + 3 = 0$ gives $x = -3$.

Axis of symmetry. Halfway between the zeros:

$$x = \frac{1 + (-3)}{2} = -1$$

Vertex. Substitute $x = -1$:

$$y = 2(-1 - 1)(-1 + 3) = 2(-2)(2) = -8$$

The vertex is $(-1, -8)$.

$y$-intercept. Substitute $x = 0$: $y = 2(0 - 1)(0 + 3) = 2(-1)(3) = -6$.

Minimum or maximum. $a = 2 \gt 0$, so the parabola opens up and the minimum value is $-8$.

Example 2: Sketching from factored form

Sketch $y = 0.5(x + 2)(x - 4)$.

Solution.

1. Zeros: $-2$ and $4$.
2. Axis of symmetry: $x = \dfrac{-2 + 4}{2} = 1$.
3. Vertex: $y = 0.5(1 + 2)(1 - 4) = 0.5(3)(-3) = -4.5$, so the vertex is $(1, -4.5)$.
4. $y$-intercept: $y = 0.5(0 + 2)(0 - 4) = 0.5(2)(-4) = -4$, so the graph passes through $(0, -4)$. Its mirror image across $x = 1$ is $(2, -4)$.
5. Since $a = 0.5 \gt 0$, the parabola opens up. Draw a smooth curve through the points, as in the graph above.

Example 3: Factoring first

Write $y = 3x^2 - 6x - 24$ in factored form. Then find the zeros and the vertex.

Solution. Take out the common factor $3$, then factor the trinomial. You need two numbers that multiply to $-8$ and add to $-2$: they’re $-4$ and $2$.

$$\begin{aligned} y &= 3x^2 - 6x - 24 \\ &= 3(x^2 - 2x - 8) \\ &= 3(x - 4)(x + 2) \end{aligned}$$

Check by expanding: $3(x^2 + 2x - 4x - 8) = 3(x^2 - 2x - 8) = 3x^2 - 6x - 24$. ✓

The zeros are $4$ and $-2$. The axis is $x = \dfrac{4 + (-2)}{2} = 1$, and

$$y = 3(1 - 4)(1 + 2) = 3(-3)(3) = -27$$

The vertex is $(1, -27)$.

Example 4: Finding the equation from the zeros and a point

A parabola has zeros at $-1$ and $3$ and passes through $(1, 8)$. Find its equation in factored form and in standard form.

Solution. Write the factored form with the zeros filled in:

$$y = a(x + 1)(x - 3)$$

Substitute $(1, 8)$:

$$\begin{aligned} 8 &= a(1 + 1)(1 - 3) \\ 8 &= a(2)(-2) \\ 8 &= -4a \\ a &= -2 \end{aligned}$$

Factored form: $y = -2(x + 1)(x - 3)$.

For standard form, expand the brackets first, then multiply by $-2$:

$$\begin{aligned} y &= -2(x^2 - 3x + x - 3) \\ &= -2(x^2 - 2x - 3) \\ &= -2x^2 + 4x + 6 \end{aligned}$$

Check: the point $(1, 8)$ is halfway between the zeros, so it’s the vertex. Since $a \lt 0$, the parabola opens down and $8$ is its maximum value, which makes sense.

Common mistakes

Getting the signs of the zeros wrong. $y = (x + 4)(x - 2)$ has zeros $-4$ and $2$, not $4$ and $-2$. Set each bracket equal to $0$ and solve.

Missing a zero from a factor like $2x - 3$. The zero is $\tfrac{3}{2}$ (solve $2x - 3 = 0$), not $3$.

Forgetting the zero from a lone $x$. $y = -3x(x - 5)$ has two zeros, $0$ and $5$. The factor $x$ counts.

Thinking $a$ changes the zeros. The value of $a$ stretches the parabola, but it doesn’t move the zeros: $y = (x - 1)(x - 5)$ and $y = 7(x - 1)(x - 5)$ both have zeros $1$ and $5$. But $a$ does change the vertex’s $y$-coordinate, so include it when you substitute.

Forgetting the common factor. For $y = 3x^2 - 6x - 24$, take out the $3$ first. Keep it in the final answer: $3(x - 4)(x + 2)$, not just $(x - 4)(x + 2)$.

Practice

1. (Warm-up) Find the zeros of each relation.

• (a) $y = (x - 6)(x + 2)$
• (b) $y = -3x(x - 5)$
• (c) $y = (2x + 1)(x - 4)$

Solution

(a) $6$ and $-2$.

(b) $0$ and $5$.

(c) $2x + 1 = 0$ gives $x = -\tfrac{1}{2}$, and $x - 4 = 0$ gives $x = 4$. The zeros are $-\tfrac{1}{2}$ and $4$.

2. (Warm-up) Find the equation of the axis of symmetry of $y = 4(x - 3)(x + 9)$.

Solution

The zeros are $3$ and $-9$:

$$x = \frac{3 + (-9)}{2} = \frac{-6}{2} = -3$$

The axis of symmetry is $x = -3$.

3. (Warm-up) Find the $y$-intercept of $y = -2(x - 1)(x - 5)$.

Solution

Substitute $x = 0$:

$$y = -2(0 - 1)(0 - 5) = -2(-1)(-5) = -10$$

The $y$-intercept is $-10$.

4. (Core) For $y = -(x - 2)(x - 8)$, find the zeros, the axis of symmetry, the vertex, the $y$-intercept, and the maximum or minimum value. Then sketch the graph.

Solution

• Zeros: $2$ and $8$.
• Axis: $x = \dfrac{2 + 8}{2} = 5$.
• Vertex: $y = -(5 - 2)(5 - 8) = -(3)(-3) = 9$, so $(5, 9)$.
• $y$-intercept: $y = -(0 - 2)(0 - 8) = -(-2)(-8) = -16$.
• $a = -1 \lt 0$, so it opens down with a maximum value of $9$.

Sketch: plot $(2, 0)$, $(8, 0)$, the vertex $(5, 9)$, and $(0, -16)$ with its mirror image $(10, -16)$, then draw a smooth curve opening down.

5. (Core) Write $y = x^2 + 4x - 21$ in factored form. Find the zeros and the vertex.

Solution

Two numbers that multiply to $-21$ and add to $4$: $7$ and $-3$.

$$y = (x + 7)(x - 3)$$

The zeros are $-7$ and $3$. The axis is $x = \dfrac{-7 + 3}{2} = -2$, and

$$y = (-2 + 7)(-2 - 3) = (5)(-5) = -25$$

The vertex is $(-2, -25)$.

6. (Core) Write $y = 2x^2 + x - 6$ in factored form. Find the zeros and the vertex.

Solution

For $ax^2 + bx + c$, look for two numbers that multiply to $ac = 2(-6) = -12$ and add to $b = 1$: they’re $4$ and $-3$. Split the middle term and group:

$$\begin{aligned} y &= 2x^2 + 4x - 3x - 6 \\ &= 2x(x + 2) - 3(x + 2) \\ &= (2x - 3)(x + 2) \end{aligned}$$

The zeros are $\tfrac{3}{2}$ and $-2$. The axis is

$$x = \frac{\tfrac{3}{2} + (-2)}{2} = \frac{-\tfrac{1}{2}}{2} = -\frac{1}{4}$$

Substitute $x = -\tfrac{1}{4}$, or $-0.25$:

$$y = \big(2(-0.25) - 3\big)(-0.25 + 2) = (-3.5)(1.75) = -6.125$$

The vertex is $(-0.25, -6.125)$.

7. (Core) A parabola has zeros at $2$ and $6$ and passes through $(4, -8)$. Find its equation in factored form, and state its vertex.

Solution

$$y = a(x - 2)(x - 6)$$

Substitute $(4, -8)$:

$$-8 = a(4 - 2)(4 - 6) = a(2)(-2) = -4a \quad\Rightarrow\quad a = 2$$

The equation is $y = 2(x - 2)(x - 6)$.

The point $(4, -8)$ is halfway between the zeros, so it is the vertex: $(4, -8)$.

8. (Challenge) A soccer ball is kicked from the ground. It follows a parabolic path, lands $40$ m away, and reaches a maximum height of $10$ m. Let $x$ be the horizontal distance from the kick (in metres) and $y$ the height (in metres).

• (a) Find an equation for the path in factored form.
• (b) How high is the ball when it is $10$ m from where it was kicked?

Solution

(a) The ball is on the ground at $x = 0$ and $x = 40$, so these are the zeros:

$$y = a(x - 0)(x - 40) = ax(x - 40)$$

The maximum is halfway, at $x = 20$, where $y = 10$:

$$10 = a(20)(20 - 40) = a(20)(-20) = -400a \quad\Rightarrow\quad a = -\frac{1}{40}$$

The equation is $y = -\tfrac{1}{40}x(x - 40)$.

(b) Substitute $x = 10$:

$$y = -\frac{1}{40}(10)(10 - 40) = -\frac{1}{40}(10)(-30) = \frac{300}{40} = 7.5$$

The ball is $7.5$ m high.

9. (Challenge) Compare these three relations. Factor each one if you can, and say how many $x$-intercepts each graph has.

• (a) $y = x^2 - 6x + 5$
• (b) $y = x^2 - 6x + 9$
• (c) $y = x^2 - 6x + 10$

Solution

(a) $y = (x - 1)(x - 5)$. Two $x$-intercepts: $1$ and $5$.

(b) $y = (x - 3)(x - 3) = (x - 3)^2$. Both factors give the same zero, so the graph has only one $x$-intercept, $3$. The vertex $(3, 0)$ sits right on the $x$-axis.

(c) No two integers multiply to $10$ and add to $-6$, so this doesn’t factor. Compare it with (b): $x^2 - 6x + 10 = (x^2 - 6x + 9) + 1 = (x - 3)^2 + 1$. It’s the graph from (b) moved up $1$, so its vertex is $(3, 1)$ and it opens up. It never reaches the $x$-axis: no $x$-intercepts, which is why it has no factors. You’ll explore this case more in zeros and the discriminant.