MPM2D Factored Form and Zeros: Notes and Practice

A soccer ball is kicked straight up from the ground. Its height in metres after $t$ seconds is modelled by $h = - 5 t (t - 4)$ . Two questions matter most: when does the ball land, and how high does it go? The equation is written as a product, and that makes the first question easy. The ball is on the ground when $h = 0$ , and a product of numbers is zero only when at least one of the numbers is zero. So either $- 5 t = 0$ , giving $t = 0$ (the moment of the kick), or $t - 4 = 0$ , giving $t = 4$ (the moment it lands). The highest point comes exactly halfway between, at $t = 2$ , where $h = - 5 (2) (2 - 4) = - 5 (2) (- 2) = 20$ m. Writing a quadratic in this factored form turns finding the zeros into reading them off.

The rule used above is the zero product property: if $A \times B = 0$ , then $A = 0$ or $B = 0$ (or both). It works only when the product equals zero: knowing $A \times B = 6$ tells you nothing about $A$ or $B$ on its own. That is why factored form is so useful for zeros, which are exactly the points where $y = 0$ .

y = a (x - r) (x - s) zeros x = r and x = s, axis x = \frac{r + s}{2}

Zeros: $x = r$ makes the first bracket zero and $x = s$ makes the second bracket zero. Each zero has the opposite sign to the number shown in its bracket: $(x + 2)$ gives $x = - 2$ .
Axis of symmetry: the zeros are mirror images, so the axis is their average, $x = \frac{r + s}{2}$ .
Vertex: substitute the axis value into the equation to get the $y$ -coordinate. This gives the maximum or minimum value.
y-intercept: substitute $x = 0$ : $y = a (0 - r) (0 - s) = a \cdot r \cdot s$ .
Direction: as in every form, $a > 0$ opens up and $a < 0$ opens down. The value of $a$ is the same as the $a$ in standard and vertex form.

Fig 1The parabola y = (x + 2)(x − 4) with zeros −2 and 4, the axis x = 1 halfway between them, vertex (1, −9) and y-intercept −8. The dashed parabola y = −½(x + 2)(x − 4) has the same zeros but a different a, so it opens down with vertex (1, 4.5).

Factor	Set it to zero	Zero
$(x - 4)$	$x - 4 = 0$	$x = 4$
$(x + 2)$	$x + 2 = 0$	$x = - 2$
$x$	$x = 0$	$x = 0$
$(2 x - 3)$	$2 x - 3 = 0$	$x = \frac{3}{2} = 1.5$
$(x - 3)^{2}$	$x - 3 = 0$	$x = 3$ only (both factors give the same zero)

Reading zeros from factors: solve factor = 0 for each one

From standard form to factored form

Everything you learned about factoring in the last unit now has a graphical meaning: the factors of a quadratic expression give the x-intercepts of its graph. To find the zeros of $y = x^{2} - 2 x - 8$ , factor it. You need two numbers that multiply to $- 8$ and add to $- 2$ : they are $- 4$ and $2$ , so $y = (x - 4) (x + 2)$ and the zeros are $4$ and $- 2$ . Always look for a common factor first: $y = 2 x^{2} - 4 x - 30 = 2 (x^{2} - 2 x - 15) = 2 (x - 5) (x + 3),$ with zeros $5$ and $- 3$ . A common factor that is only a number, like the 2 here, does not create a zero, but a common factor containing $x$ does: $y = 2 x^{2} + 6 x = 2 x (x + 3)$ has zeros $0$ and $- 3$ . A difference of squares works too: $y = x^{2} - 9 = (x - 3) (x + 3)$ has zeros $3$ and $- 3$ .

One zero, or no factoring at all

If both factors are the same, as in

y = (x - 3)^{2}

, the parabola has only one zero and its vertex

(3, 0)

sits on the x-axis. Some quadratics do not factor using whole numbers.

y = x^{2} + 4

has no zeros at all (its lowest point is

(0, 4)

), so it has no factored form with real numbers.

y = x^{2} - 2

does cross the x-axis, but at

x \approx - 1.41

and

x \approx 1.41

, which are not whole numbers. For parabolas like these, use vertex form or the methods in the next unit.

Sketching from factored form

Factor fully if the equation is in standard form, then plot the zeros $(r, 0)$ and $(s, 0)$ .
Draw the axis of symmetry $x = \frac{r + s}{2}$ halfway between the zeros.
Substitute the axis value to find the vertex, and plot it.
Plot the $y$ -intercept $(0, a \cdot r \cdot s)$ and its mirror point on the other side of the axis.
Draw a smooth curve through the points, opening up if $a > 0$ and down if $a < 0$ .

VideoUse the zeros from factored form, the axis of symmetry halfway between them, and the vertex to sketch a parabola.Watch on YouTube

Building an equation from its zeros

Two zeros do not fix a parabola. The parabolas $y = (x + 2) (x - 4)$ , $y = 3 (x + 2) (x - 4)$ and $y = - \frac{1}{2} (x + 2) (x - 4)$ all have zeros $- 2$ and $4$ , but different shapes. To find the exact equation, write $y = a (x - r) (x - s)$ with the zeros filled in, substitute one more point on the parabola, and solve for $a$ . A good extra point is often the $y$ -intercept or the vertex, since it is usually given or easy to read from a graph.

Loading interactive…

Type a, r and s in the factored form panel, or drag the two zeros on the graph. The axis of symmetry stays halfway between the zeros, and the vertex, the standard form and the vertex form update with them.

Factored Form and Zeros

Key things to remember

From standard form to factored form

Sketching from factored form

Building an equation from its zeros

Worked examples

Where marks get lost

Flashcards 8

Practice