Find Zeros of Polynomials

  

How to find the zeros of polynomials using factoring, division of polynomials and the rational root theorem
. Grade 12 maths questions are presented along with detailed solutions and graphical interpretations.

Question 1

Polynomial $p$ is defined by $p(x) = x^3+5x^2-2x-24 \;$ has a zero at$\; x = 2$. Factor $p$ completely and find its zeros.

solution

$p(x)$ has a zero at $x = 2$ and therefore $\; x - 2$ is a factor of $p(x)$. Divide $p(x)$ by $x - 2$

$\dfrac{p(x)}{x - 2} = \dfrac {x^3 + 5 x^2 - 2 x - 24}{x-2} = x^2 + 7 x + 12$

Using the division above, $p(x)$ may now be written in factored form as follows:
$p(x) = (x - 2)(x^2 + 7 x + 12)$
Factor the quadratic expression $x^2 + 7 x + 12$.
$p(x) = (x - 2)(x + 3)(x + 4)$
The zeros are found by solving the equation.
$p(x) = (x - 2)(x + 3)(x + 4) = 0$
For p(x) to be equal to zero, we need to have
$x - 2 = 0$, or $x + 3 = 0$ , or $x + 4 = 0$
Solve each of the above equations to obtain the zeros of $p(x)$.
$x = 2$, $x = - 3$ and $x = - 4$.

Question 2

The polynomial $p(x)=3x^4+5x^3-17x^2-25x+10$ has irrational zeros at $x = \pm \sqrt5$. Find the other zeros.

solution

Zeros at $x = \pm \sqrt5$ correspond to the factors.
$x - \sqrt 5$ and $x + \sqrt 5$
Hence polynomial $p(x)$ may be written as
$p(x) = (x - √5)(x + √5) Q(x) = (x^2 - 5)Q(x)$
Find $Q(x)$ using long division of polynomials
$Q(x) = \dfrac{p(x)}{x^2 - 5}$

$\quad = \dfrac{3 x^4 + 5 x^3 - 17 x^2 - 25 x + 10}{x^2 - 5}$

$\quad = 3 x^2 + 5 x - 2$
Factor $Q(x) = 3 x^2 + 5 x - 2$
$Q(x) = 3 x^2 + 5 x - 2 = (3x - 1)(x + 2)$
Factor $p(x)$ completely
$p(x) = (x - \sqrt 5)(x + \sqrt 5)(3x - 1)(x + 2)$
Set each of the factors of p(x) to zero to find the zeros.
$x = \pm \sqrt 5 , x = \dfrac{1}{3} , x = - 2$

Question 3

Polynomial $p$ is given by $p(x) = x^4 - 2x^3 - 2x^2 + 6x - 3$

a) Show that $x = 1$ is a zero of multiplicity $2$.

b) Find all zeros of $p$.

c) Sketch a possible graph for $p$.

solution

a)
If $x = 1$ is a zero of multiplicity $2$, then $(x - 1)^2$ is a factor of $p(x)$ and a division of $p(x)$ by $(x - 1)^2$ must give a remainder equal to $0$. A long division gives

$\dfrac{p(x)}{(x - 1)^2} = \dfrac{x^4 - 2x^3 - 2x^2 + 6x - 3}{(x - 1)^2} = x^2 - 3$

The remainder in the division of $p(x)$ by $(x - 1)^2$ is equal to $0$ and therefore $x = 1$ is a zero of multiplicity $2$.
b)
Using the division above, p(x) may now be written in factored form as follows
$p(x) = (x - 1)^2(x^2 - 3)$
Factor the quadratic expression $x^2 - 3. \( p(x) = (x - 1)^2 (x - \sqrt 3) (x + \sqrt 3)$
The zeros are found by solving the equation.
$p(x) = (x - 1)^2 (x - \sqrt 3) (x + \sqrt 3) = 0$
For $p(x)$ to be equal to zero, we need to have
$(x - 1)^2 = 0$ , or $x - \sqrt 3 = 0$ , or $x + \sqrt 3 = 0$
Solve each of the above equations to obtain the zeros of p(x).
$x = 1$ (multiplicity $2$ ) , $x = \sqrt 3$ and $x = - \sqrt 3$
c)
With the help of the factored form of $p(x)$ and its zeros found above, we now make a table of signs.

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We use the zeros of p(x) which graphically are shown as x intercepts, the table of signs and the y intercept (0 , -3) to complete the graph as shown below.

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Question 4

Use the Rational Zeros Theorem to determine all rational zeros of the polynomial $p(x) = 6x^3-13x^2+x+2$.

solution

Rational Zeros Theorem: If $p(x)$ is a polynomial with integer coefficients and if $\dfrac{m}{n}$ (in lower terms) is a zero of $p(x)$, then $m$ is a factor of the constant term $2$ of $p(x)$ and n is a factor of the leading $6$ coefficient of $p(x)$.
Find factors of $2$ and $6$.

factors of $2$ are : $\quad \pm 1$ , $\pm 2$
factors of $6$ are: $\quad \pm 1 , \pm 2 , \pm 3 , \pm 6$
possible zeros: divide factors of $2$ by factors of $6$: $\quad \pm 1 , \pm \dfrac{1}{2} , \pm \dfrac{1}{3} , \pm \dfrac{1}{6} , \pm 2 , \pm \dfrac{2}{3}$
Because of the large list of possible zeros, we graph the polynomial and guess the zeros from the location of the $x$ intercepts. Below is the graph of the the given polynomial $p(x)$ and we can easily see that the zeros are close to $- \dfrac{1}{3}$, $\dfrac{1}{2}$ and $2$.

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We now calculate $p(-\dfrac{1}{3}), p(\dfrac{1}{2})$ and $p(2)$ to finally check if these are the exact zeros of $p(x)$.

$p \left(-\dfrac{1}{3} \right) = 6\left(-\dfrac{1}{3} \right)^3 - 13 \left (-\dfrac{1}{3} \right)^2 + \left (-\dfrac{1}{3} \right) + 2 = 0$

$p \left(\dfrac{1}{2} \right) = 6 \left (\dfrac{1}{2} \right)^3 - 13 \left (\dfrac{1}{2} \right)^2 + \left (\dfrac{1}{2} \right) + 2 = 0$

$p(2) = 6(2)^3 - 13(2)^2 + (2) + 2 = 0$
We have used the rational zeros theorem and the graph of the given polynomial to determine the $3$ zeros of the given polynomial which are $-\dfrac{1}{3}, \dfrac{1}{2}$ and $2$.