2.2-Zeroes of Polynomials

2.2-Zeroes of Polynomials Important Formulae

You are currently studying
Grade 9 → Math → Polynomials → 2.2-Zeroes of Polynomials

After successful completion of this topic, you should be able to:

  • Calculate zeros of a polynomial.

2.2 - Zeroes of Polynomials
Definition of Zeroes of Polynomials

A zero of a polynomial is a value of $x$ for which the polynomial evaluates to zero. In other words, if $p(x)$ is a polynomial, then $c$ is a zero of $p(x)$ if: $$p(c) = 0$$ This means that when we substitute $c$ into the polynomial, the result is zero.

Types of Polynomials
Polynomials can be classified based on their degrees:
  1. Linear Polynomials: These are polynomials of degree 1. Example: $p(x) = ax + b$. - A linear polynomial has exactly one zero.
  2. Quadratic Polynomials: These are polynomials of degree 2. Example: $p(x) = ax^2 + bx + c$. - A quadratic polynomial can have two, one, or no real zeroes depending on the discriminant $D = b^2 - 4ac$. - If $D > 0$, there are two distinct real zeroes. - If $D = 0$, there is one real zero (repeated). - If $D < 0$, there are no real zeroes.
  3. Cubic Polynomials: These are polynomials of degree 3. Example: $p(x) = ax^3 + bx^2 + cx + d$. - A cubic polynomial can have one real zero or three real zeroes.
  4. Higher-Degree Polynomials: Polynomials of degree $n$ can have up to $n$ real zeroes.
Finding Zeroes of Polynomials
To find the zeroes of a polynomial, we set the polynomial equal to zero and solve for $x$.
  1. Factoring Method: This involves expressing the polynomial as a product of its factors. For example, to find the zeroes of $p(x) = x^2 - 5x + 6$, we can factor it as: $$p(x) = (x - 2)(x - 3)$$ Setting $p(x) = 0$, we get the zeroes $x = 2$ and $x = 3$.
  2. Quadratic Formula: For a quadratic polynomial $ax^2 + bx + c$, the zeroes can be found using: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
  3. Graphical Method: The zeroes of a polynomial can also be identified graphically where the polynomial intersects the x-axis.


Googolplexian1221, CC0, via Wikimedia Commons

Find the value of the polynomial 5x – 4x$^2$ + 3 at:

(i) x=0
(ii) x=–1
(iii) x=2

Solution:
(i) 3
(ii) -6
(iii) -3

Value of the Polynomial 5x – 4x$^2$ + 3

(i) For x=0:
$5(0) – 4(0)^2 + 3 = 3$

(ii) For x=–1:
$5(–1) – 4(–1)^2 + 3 = –5 – 4 + 3 = –6$

(iii) For x=2:
$5(2) – 4(2)^2 + 3 = 10 – 16 + 3 = –3$

Suggest improvements/Report errors

Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y) = y$^2$ –y+1
(ii) p(t) = 2+t+2t$^2$ –t$^3$
(iii) p(x) = x$^3$
(iv) p(x) = (x–1)(x+1)

Solution:

Polynomial Evaluations

(i) p(0) = 1, p(1) = 1, p(2) = 3
(ii) p(0) = 2, p(1) = 4, p(2) = 4
(iii) p(0) = 0, p(1) = 1, p(2) = 8
(iv) p(0) = -1, p(1) = 0, p(2) = 3

Suggest improvements/Report errors

Verify whether the following are zeroes of the polynomial, indicated against them.

  1. p(x)=3x+1, x= $\dfrac{-1}{3}$
  2. p(x)=5x–$\pi$, x= $\dfrac{4}{5}$
  3. p(x)=x$^2$ –1, x=1,–1
  4. p(x)=(x+1)(x–2), x=–1, 2
  5. p(x)=x$^2$, x=0
  6. p(x)=lx+m, x= $–\dfrac{m}{l}$
  7. p(x)=3x$^2$ –1, x= $\dfrac{−1}{\sqrt{3}}$, $\dfrac{2}{\sqrt{3}}$
  8. p(x)=2x+1, x= $\dfrac{1}{2}$

Solution:
(i) Yes
(ii) No
(iii) Yes
(iv) Yes
(v) Yes
(vi) Yes

Suggest improvements/Report errors

Find the zero of the polynomial in each of the following cases:
(i) p(x) = x+5
(ii) p(x) = x–5
(iii) p(x) = 2x+5
(iv) p(x) = 3x–2
(v) p(x) = 3x
(vi) p(x) = ax, a $\neq$ 0
(vii) p(x) =cx + d, c $\neq$ 0, c, d are real numbers.

Solution:

Finding Zeros of Polynomials

(i) Set $p(x) = 0$: $x + 5 = 0 \implies x = -5$.
(ii) Set $p(x) = 0$: $x - 5 = 0 \implies x = 5$.
(iii) Set $p(x) = 0$: $2x + 5 = 0 \implies x = -\frac{5}{2}$.
(iv) Set $p(x) = 0$: $3x - 2 = 0 \implies x = \frac{2}{3}$.
(v) Set $p(x) = 0$: $3x = 0 \implies x = 0$.
(vi) Set $p(x) = 0$: $ax = 0 \implies x = 0$.
(vii) Set $p(x) = 0$: $cx + d = 0 \implies x = -\frac{d}{c}$.

Suggest improvements/Report errors