# 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.

## Questions with Solutions

### 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
$$p(x) = (x - \sqrt {5})(x + \sqrt {5}) Q(x) = (x^2 - 5)Q(x)$$
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 $$2$$.