Sales Toll Free No: 1-855-666-7446
• Math /
• Algebra 2 /
• Exploring Polynomials and Radical Expression

# Exploring Polynomials and Radical Expression

Top
 Sub Topics An expression which consists of constant, variables and exponent values is known as polynomial expression. The polynomial expression is combined with the addition, subtraction, and multiplication.Like $5x^{2} + 9x + 12$ = $0$;This given expression is a polynomial expression and one exponent value is present in this expression.The expression which contains a Square root values is known as radical expression.For example: $\sqrt{6},\ 3 \sqrt{x + 2},\ 8 \sqrt{16}$ and $6 + 3 \sqrt{12}$; these all are radical expressions.The symbol $\sqrt{}$ is used to represents the square root or $n$ th roots of the expression.The other definition, an expression which contains the radical expression with the variable in the radicands is also known as radical equation.Here we will be exploring Polynomials and radical expression.Suppose we have a radical expression $\sqrt{\frac{x^{4}-17}{y-5}}$Where the value of ‘$x$’ is $3$ and the value of ‘$y$’ is $9$ and we have to solve the radical expression;For solving this expression we follow some steps:The given expression is $\sqrt{\frac{x^{4}-17}{y-5}}$And value of ‘$x$’ and ‘$y$’ is $3$ and $9$;Step 1: First substitute the values of ‘$x$’ and ‘$y$’ in the given radical expression.$\Rightarrow$ $\sqrt{\frac{(3)^{4} - 17}{9-5}}$Step 2: Then we find the positive square root.$\Rightarrow$ $\sqrt{\frac{64}{4}}$Step 3: Then solve the expression.$\Rightarrow$ $\sqrt{16}$$\Rightarrow$ $4$;Where $x$ = $3$ and $y$ = $9$Now we will see how to solve polynomial expression;Suppose we have $3x^{2} -5x + 3$ at $x - 4$;For solving this expression we have to follow some steps:Step 1: First we have a polynomial expression.Step 2: Then put the value of ‘$x$’ as $-4$.$\Rightarrow$ $3x^{2} -5x + 3$;On putting the value of $x$ = $-4$ we get:$\Rightarrow$ $3(-4)^{2} -5(-4) + 3$;Step 3: Then we solve the expression:$\Rightarrow$ $3 \times 16 + 20 + 3$;$\Rightarrow$ $48 + 23$;$\Rightarrow$ $71$;

## Analyzing Graphs of Quadratic Equations

Quadratic equation can be defined as polynomial equation with highest degree $2$. Quadratic equation can also be called as a second degree equation. Consider the following function $f (X)$ = $J X^{2} + KX + L$, Since highest degree of variable '$X$' is two $(2)$ hence it is called as Quadratic Equation. $J, K$ and $L$ are quadratic coefficients. Lets see how to Graph Quadratic Equation.

Domain of the above function $f (X)$ is the Set of all Real Numbers. $y$-intercept of graph of $f (X)$ is given by $f (0)$ = $c$ and $x$-intercepts are obtained by solving the equation $JX^{2} + KX + L$ = $0$ using the Quadratic Formula.

Lets understand the process of analyzing graphs of quadratic Functions. Range of quadratic function $f (X)$ can be found as shown below:

First, $f (X)$ = $J (X - h)^{2} + C$,

$f (X)$ = $JX^{2} - 2JhX + Jh^{2} + C$

Two conditions must be followed $-2Jh$ = $K$ (first equation) and $Jh^{2} + C$ = $L$ (second equation). From first equation, $h$ = $\frac{-K}{2J}$. Using second equation,

$C$ = $\frac{L - K^{2}}{4J}$

The term $(X - h)^{2}$ is either positive or zero, hence $(X - h)^{2} \geq 0$.

If $J > 0$, multiply both sides of inequality above by $J$

$J(X - h)^{2} \geq 0$,

Add $C$ to both sides of inequality

$J(X - h)^{2} + C \geq C$,

Thus $f (X) \geq C$

Thus range of $f (X)$ is given by

$(C, + \infty)$. Since minimum value of $f (X)$ is $C$ and it can be concluded that range of $f (X)$ can also be given by $(- \infty, C)$.

From above discussion, it is clear that graph of quadratic function is called as Parabola and Point with coordinates $(h, C)$ is called vertex of parabola.

Consider the equation $f (x)$ = $2x^{2} + 2 x - 4$, for which graph looks like this:

## Solving Quadratic Equations by Factoring

Quadratic equation is an equation whose highest degree is two. Degree $2$ indicates that highest power in equation is two.

Let’s discuss process of solving Quadratic Equation by factoring. Let’s consider the following quadratic equation to understand how to solve quadratic equations by factoring?

$A\ y^{2} + B\ y + C$ = $0$,

Since maximum degree of variable '$y$' is two $(2)$, hence this type of equation is referred to quadratic equation. $A, B$ and $C$ are called quadratic coefficients. There are two methods to solve a quadratic equation. One is direct Factorization and second is using Quadratic Formula.

Above equation can be solved using quadratic formula which is given as:

$y_{0}$ = $\frac{-B \pm \sqrt{b^{2} - 4AC}}{2A}$

This is the standard formula for factorizing quadratic equation.

Let’s take following example:

Suppose we are given $P^{2} + 9P + 20$ = $0$ then what is the solution of this equation? There are two types of factorization methods. One method includes direct factorization. In direct factorization, we multiply first $(P^{2})$ and last $(20)$ terms which gives a new value $(20P^{2})$. Then distribute the middle value $(9P)$ in such a way that its multiplication gives $(20P^{2})$ and addition or subtraction gives middle value. Another method is use of quadratic formula.

Using the first method, we get following solution:

$P^{2} + 9P + 20$ = $P^{2} + 5P + 4P + 20$ = $0$,

$P (P + 5) + 4 (P + 5)$ = $0$,

$(P + 5) (P + 4)$ = $0$,

Then solution will be,

$P$ = $-5$ or $P$ = $-4$,

Standard Quadratic Equation is written in form $ax^{2} + b x + c$ = $0$, but in case of quadratic inequalities expression is written in as,

1) $ax^{2} + b\ x + c > 0$, example: $x^{2} + 4x > 5$,

2) $ax^{2} + b\ x + c < 0$, example: $8x^{2} < 29$,

3) $ax^{2} + b\ x + c < 0$, example: $6 \geq x^{2} – x$,

4) $ax^{2} + b\ x + c \leq 0$, example: $4y^{2} + 1$ = $8y$

There are some steps for solving quadratic inequalities :

Step 1: First of all we have to move all terms to one side.

Step 2: Factorize the inequality.

Step 3: After this find the roots of this quadratic equation.

Step 4: We can graph quadratic inequalities by plotting coordinates on graph and then shading appropriate area according to inequality sign.

Let’s take an example: $x^{2} + 4x < 5$,

Solution:

First of all we will move $5$ to left side

$x^{2} + 4x – 5 < 0$, now solve this equation and find the root, so, $x^{2} + 5x - x - 5 < 0 \Rightarrow x (x + 5) - 2 (x + 5) < 0$ ,

$\Rightarrow\ (x - 2) (x + 5) < 0$, so $x$ = $2,\ - 5$.

Let’s take the value $x$ = $0$ then $0 – 0 -5 < 0$

Now $x$ = $1, 1 + 4 -5 < 0$

Now $x$ = $-4 \Rightarrow 16 – 16 -5 < 0$

Now $x$ = $-3 \Rightarrow 9 – 12 – 5 < 0$

So it is proved that $x^{2} + 4x – 5 < 0$ in interval $(2 -5)$.

Quadratic equation is written as $ax^{2} + bx + c$ = $0$, from this equation we can calculate Quadratic Formulas and Discriminant. Quadratic

formula is given as: $\frac{(-b\ \mp\ \sqrt{(b^{2}\ –\ 4ac)})}{2a}$, this formula is called, this formula gives us two solutions,

$x_{1}$ = $\frac{(-b\ +\ \sqrt{(b^{2}\ –\ 4ac)})}{2a}$ and $x_{2}$ = $\frac{(-b\ -\ \sqrt{(b^{2}\ –\ 4ac)})}{2a}$.

Similarly we can find the Discriminant, discriminant provides the information about nature of roots and solution of any Quadratic Equation. Discriminant formula is given by:

Discriminant = $b^{2}\ –\ 4ac$.

Suppose a quadratic equation $x^{2}\ +\ 4x\ +\ 4$ = $0$, here $a$ = $1,\ b$ = $4,\ c$ = $4$ so discriminant is $4^{2}\ –\ 4ac\ \Rightarrow\ 16\ –\ 4\ .\ 1\ .\ 4\ \Rightarrow\ 16\ -\ 16$ = $0$.
We can also find the nature of solution using this discriminant .

Case 1: If value of $b^{2} – 4 a c > 0$ means discriminant is positive discriminant, in this case two real solutions are generated, if discriminant is a perfect Square then roots are rational, otherwise they are irrational.

Case 2: If value of discriminant is zero, $b^{2} – 4 a c$ = $0$, then there is only one real solution.

Case 3: If value of discriminant is negative, $b^{2} – 4 a c < 0$, then in this case there is no real solution, there will be two imaginary solutions.
Example:

Calculate the discriminant, nature and number of solutions of equation $y$ = $x^{2} + 4x + 5$, in this equation $a$ = $1,\ b$ = $4,\ c$ = $5$, so discriminant = $b^{2} – 4 a c?$

Solution:

Put value of $a, b, c$ in this formula, then $(42 – 4 .1.5) \Rightarrow\ 16 – 20$ = $-4$.

Here discriminant is negative, so there are no real solution for this quadratic equation, solutions are imaginary. So using Quadratic Formula for

$y$ = $x^{2} + 4x + 5,$

$X_{1}$ = $(-4) +-$ $\frac{\sqrt{(4^{2} – 4 .1.5)})}{2.1}$

= $\frac{(4 - \sqrt{(-4)})}{2}$,

= $\frac{(4 + 2 i)}{2}$, $\frac{(4 - 2 i)}{2}$,

= $2 + i, 2 - i$

## Solving Quadratic Equations by Graphing

An equation with highest degree of $2$ is called as Quadratic Equation. We can say that an equation in which highest power is a Square is known as quadratic equation. If highest power is more than $2$ then it is not a quadratic equation. It can be written as: $ax^{2} + bx + c$ = $0$ and Quadratic Formula is given by: