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

Using the division above, $p(x)$ may now be written in factored form as follows: $$p(x)=(x-2)(x^2+7x+12)$$

Factor the quadratic expression $x^2+7x+12$. $$x^2+7x+12=(x+3)(x+4)$$ and substitute in $P(x)$ $$p(x)=(x-2)(x+3)(x+4)$$

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 $$x-2=0\text{or}x+3=0\text{or}x+4=0$$

Solve each of the above equations to obtain the zeros of $p(x)$. $$x=2,x=-3,x=-4$$

Question 2

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

solution

Zeros at $x=\pm \sqrt{5}$ correspond to the factors. $$x-\sqrt{5}\text{and}x+\sqrt{5}$$ 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 $$Q(x)={\displaystyle \frac{p(x)}{x^2-5}}$$ $$={\displaystyle \frac{3x^4+5x^3-17x^2-25x+10}{x^2-5}}$$ $$=3x^2+5x-2$$

Factor $Q(x)=3x^2+5x-2$ $$Q(x)=3x^2+5x-2=(3x-1)(x+2)$$

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 to obtain all the zeros. $$x=\pm \sqrt{5},x={\displaystyle \frac{1}{3}},x=-2$$

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

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 $$(x-1{)}^2=0,x-\sqrt{3}=0,\text{or}x+\sqrt{3}=0$$

Solve each of the above equations to obtain the zeros of $p(x)$.

$$x=1\text{ (multiplicity }2),x=\sqrt{3},x=-\sqrt{3}$$

.

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.

.

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 ${\displaystyle \frac{m}{n}}$ (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 : $\pm 1$ , $\pm 2$

factors of $6$ are: $\pm 1,\pm 2,\pm 3,\pm 6$

possible zeros: divide factors of $2$ by factors of $6$: $\pm 1,\pm {\displaystyle \frac{1}{2}},\pm {\displaystyle \frac{1}{3}},\pm {\displaystyle \frac{1}{6}},\pm 2,\pm {\displaystyle \frac{2}{3}}$

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 $-{\displaystyle \frac{1}{3}}$, ${\displaystyle \frac{1}{2}}$ and $2$. .

We now calculate $p(-{\displaystyle \frac{1}{3}}),p({\displaystyle \frac{1}{2}})$ and $p(2)$ to finally check if these are the exact zeros of $p(x)$. $$p(-{\displaystyle \frac{1}{3}})=6{(-{\displaystyle \frac{1}{3}})}^3-13{(-{\displaystyle \frac{1}{3}})}^2+(-{\displaystyle \frac{1}{3}})+2=0$$ $$p\left({\displaystyle \frac{1}{2}}\right)=6{\left({\displaystyle \frac{1}{2}}\right)}^3-13{\left({\displaystyle \frac{1}{2}}\right)}^2+\left({\displaystyle \frac{1}{2}}\right)+2=0$$ $$p(2)=6(2{)}^3-13(2{)}^2+(2)+2=0$$ 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 $-{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}}$ and $2$.

More References and links

• Introduction to Polynomials
• Factor Polynomials
• Long Division of Polynomials
• Factor a Polynomial Using Rational Root and Factor theorems
• Factoring of Special Polynomials
• Factor Polynomials by Grouping
• Find Zeros of Polynomial Functions
• High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers
• Home Page