Value, zeros and roots of a polynomial

A poly­no­mial is an expres­sion; put an equals sign beside it and it becomes an equa­tion that we can solve. This les­son con­nects the two ideas. We name equa­tions by their degree, learn to find the value of a poly­no­mial at a given num­ber, and then meet the most impor­tant num­bers attached to any poly­no­mial: its zeros. Find­ing zeros is the same job as find­ing the roots of an equa­tion, and we fin­ish by solv­ing qua­dratic equa­tions by split­ting the mid­dle term.

Equa­tions#

If two expres­sions are con­nected by the equal­ity sign ($=$), the result is an equa­tion. Equa­tions are named by their degree, exactly as poly­no­mi­als are.

Lin­ear equa­tion (degree $1$)

• In one vari­able $x$, the gen­eral form is $ax + b = 0$, where $a \neq 0$. Exam­ple: $2x + 3 = 0$.
• In two vari­ables $x$ and $y$, the gen­eral form is $ax + by + c = 0$, where $a \neq 0$ and $b \neq 0$. Exam­ple: $2x + 3y - 4 = 0$.

Qua­dratic equa­tion (degree $2$): $ax^2 + bx + c = 0$, where $a \neq 0$. Exam­ples: $2x^2 + 3x + 4 = 0$, $\ x^2 + 5 = 0$, $\ x^2 = 0$.

Cubic equa­tion (degree $3$): $ax^3 + bx^2 + cx + d = 0$, where $a \neq 0$. Exam­ples: $2x^3 + 5x^2 + 6x + 7 = 0$, $\ x^3 + 5 = 0$.

Bi-qua­dratic equa­tion (degree $4$): $ax^4 + bx^3 + cx^2 + dx + e = 0$, where $a \neq 0$. Exam­ples: $3x^4 + 5x^3 + 6x^2 + 7x + 8 = 0$, $\ x^4 + 3 = 0$.

Value of a poly­no­mial#

The value of a poly­no­mial $p(x)$ at $x = \alpha$ is found by sub­sti­tut­ing $\alpha$ for $x$ every­where in the poly­no­mial. It is writ­ten $p(\alpha)$.

Exam­ple 1#

Find the value of $p(x) = 8x^2 - 3x + 7$ at $x = -2$.

$$\begin{aligned} p(-2) &= 8(-2)^2 - 3(-2) + 7 \\ &= 8(4) + 6 + 7 \\ &= 32 + 13 \\ &= 45 \end{aligned}$$

Put brack­ets round the neg­a­tive num­ber every time you sub­sti­tute it. With­out them, $-2^2$ is eas­ily mis­read as $-4$ instead of $(-2)^2 = 4$.

Zero of a poly­no­mial#

A num­ber $\alpha$ is a zero of the poly­no­mial $p(x)$ if $p(\alpha) = 0$. In words: a zero is a value of $x$ that makes the poly­no­mial equal to noth­ing.

Exam­ple 2#

Is $2$ a zero of the poly­no­mial $2x - 4$?

Let $p(x) = 2x - 4$. Then

$$p(2) = 2(2) - 4 = 4 - 4 = 0$$

So yes, $2$ is a zero of $2x - 4$.

We could also have solved $2x - 4 = 0$ directly: $2x = 4$, so $x = 2$. That sec­ond way is how we find zeros when we are not given them.

To find the zeros of a poly­no­mial $p(x)$, solve the equa­tion $p(x) = 0$.

Exam­ple 3#

Find the zero of the poly­no­mial $2x + 3$.

$$\displaystyle 2x + 3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}$$

In gen­eral, the lin­ear poly­no­mial $ax + b$ (with $a \neq 0$) has exactly one zero, $\displaystyle x = -\frac{b}{a}$.

Root of an equa­tion#

A num­ber $\alpha$ is called a root of the equa­tion $p(x) = 0$ if $p(\alpha) = 0$.

So "zero of the poly­no­mial $p(x)$" and "root of the equa­tion $p(x) = 0$" describe the same num­ber. We say zero when we talk about the poly­no­mial and root when we talk about the equa­tion.

Exam­ple 4#

Check whether $2$ and $3$ are roots of $x^2 - 5x + 6 = 0$.

Let $p(x) = x^2 - 5x + 6$.

$$p(2) = 4 - 10 + 6 = 10 - 10 = 0$$

$$p(3) = 9 - 15 + 6 = 15 - 15 = 0$$

Both give $0$, so $2$ and $3$ are roots of $x^2 - 5x + 6 = 0$.

Solv­ing a qua­dratic by split­ting the mid­dle term#

Exam­ple 4 checked num­bers we were given. To find the roots our­selves we fac­torise. For $x^2 + bx + c$, look for two num­bers whose prod­uct is $c$ and whose sum is $b$, and use them to split the mid­dle term.

Exam­ple 5#

Solve $x^2 - 5x + 6 = 0$.

We need two num­bers with prod­uct $6$ and sum $-5$. The fac­tor pairs of $6$ are $1 \times 6$ and $2 \times 3$, and since the sum is neg­a­tive we take $(-2) \times (-3)$.

$$\begin{aligned} x^2 - 5x + 6 &= 0 \\ x^2 - 2x - 3x + 6 &= 0 \\ x(x - 2) - 3(x - 2) &= 0 \\ (x - 2)(x - 3) &= 0 \end{aligned}$$

A prod­uct is zero only when one of its fac­tors is zero, so $x - 2 = 0$ or $x - 3 = 0$, giv­ing $x = 2$ or $x = 3$. These are the same roots we checked in Exam­ple 4.

Exam­ple 6#

Solve $x^2 - 7x + 6 = 0$.

Prod­uct $6$, sum $-7$: the pair is $-1$ and $-6$.

$$\begin{aligned} x^2 - x - 6x + 6 &= 0 \\ x(x - 1) - 6(x - 1) &= 0 \\ (x - 1)(x - 6) &= 0 \end{aligned}$$

So $x = 1$ or $x = 6$.

Exam­ple 7#

Solve $x^2 - 15x + 56 = 0$.

Prod­uct $56$, sum $-15$. The fac­tor pairs of $56$ include $7 \times 8$, and $7 + 8 = 15$, so the pair is $-7$ and $-8$.

$$\begin{aligned} x^2 - 7x - 8x + 56 &= 0 \\ x(x - 7) - 8(x - 7) &= 0 \\ (x - 7)(x - 8) &= 0 \end{aligned}$$

So $x = 7$ or $x = 8$.

Exam­ple 8#

If $x = 0$ and $x = -1$ are zeros of $f(x) = 2x^3 - 3x^2 + ax + b$, find $a$ and $b$.

A zero makes the poly­no­mial equal to $0$.

$$f(0) = 0 \implies b = 0$$

$$f(-1) = 0 \implies -2 - 3 - a + b = 0 \implies -5 - a + 0 = 0 \implies a = -5$$

So $a = -5$ and $b = 0$.

Prac­tice#

1. Name each equa­tion as lin­ear, qua­dratic, cubic or bi-qua­dratic: (a) $3x - 4 = 0$ (b) $x^3 + 2x = 5$ (c) $2x^2 - 5 = 0$ (d) $x^4 - 1 = 0$ (e) $2x + 3y = 6$
2. If $f(x) = 2x^3 - 13x^2 + 17x + 12$, find $f(2)$, $f(-3)$ and $f(0)$.
3. Find the zero of each poly­no­mial: (a) $x - 5$ (b) $2x + 5$ (c) $3x - 2$ (d) $2x$
4. Is $\displaystyle x = \frac{1}{2}$ a zero of $2x + 1$? Give the value of the poly­no­mial there.
5. Ver­ify that $1$ and $-1$ are both zeros of $x^2 - 1$.
6. Show that $x = 1$ is a root of $2x^3 - 3x^2 + 7x - 6 = 0$.
7. If $x = 2$ is a root of $2x^2 - 3x + 7a = 0$, find $a$.
8. Solve $x^2 + 9x + 20 = 0$.
9. Solve $x^2 - 3x - 40 = 0$.
10. Solve $2x^2 - 3x - 35 = 0$.
11. Solve $x^2 - 38x + 357 = 0$.
12. If $\displaystyle x = \frac{4}{3}$ is a root of $6x^3 - 11x^2 + kx - 20 = 0$, find $k$.

Answers#