# 9th Class Mathematics Polynomials Types of Polynomials

Types of Polynomials

Category : 9th Class

### Types of Polynomials

We can classify polynomials by two methods

(a) According to degree

(b) According to terms

According to Degree

According to degree, polynomial can be classified into following types:

•   Linear Polynomial
•   Cubic Polynomial
•   Biquadratic Polynomial and so on.

Linear Polynomial

A polynomial in which highest power of the variable is 1. The general form of linear polynomial is $ax+b,a\ne 0$.

(i) $x+y+z=9$ is a linear polynomial

(ii) $3x+8$ is also a linear polynomial

(ii) ${{(x-2)}^{2}}-{{(x-3)}^{2}}$ is a linear polynomial [Because${{(x-2)}^{2}}-{{(x-3)}^{2}}=2x-2$]

A polynomial in which the highest power of a variable is 2. The general form of quadratic polynomial is $a{{x}^{3}}+b{{x}^{2}}+c;\,a\ne 0$.

(i) ${{x}^{2}}+4x+3$ is a quadratic polynomial

(ii) ${{(x+2)}^{2}}+{{(x-2)}^{2}}$ is a quadratic polynomial

Cubic Polynomial

A polynomial in which the highest power of variable is 3. The general form of cubic polynomial is $a{{x}^{3}}+b{{x}^{2}}+cx+d$, where $a\ne 0$.

(i) $4{{x}^{3}}+3{{x}^{2}}+4x+5$ is a cubic polynomial

(ii) ${{(4x+3)}^{3}}$ is a cubic polynomial

A polynomial in which highest power of variable is 4. The general form of biquadratic polynomial is $a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+5$.

(i) $19{{x}^{4}}+16{{x}^{3}}+13{{x}^{2}}+14x+5$ is a biquadratic polynomial

(ii) ${{(x+2)}^{4}}$ is a biquadratic polynomial

According to Number of Terms

According to number of terms, polynomials can be classified into following types.

•   Monomials
•   Binomials
•   Trinomials

Monomials

A polynomial which has only one term is known as monomials.

(i) ${{x}^{2}}yz$ is a monomial

(ii) $\frac{1}{3}xyz$ is a monomial

Binomials

A polynomial which has only two terms is known as binomials.

(i) $4{{x}^{2}}y+4z{{y}^{2}}$ is a binomial

Trinomials

A polynomial which has only three terms is known as trinomials.

(i) $4{{x}^{3}}+4{{x}^{2}}+3x$ is a trinomial

Remainder Theorem

When $P(x)$ is divided by $(x-a)$,$P(a)$ is the remainder, where $P(x)$ is a polynomial of degree n > 1 and "a" is any real number.

Factor Theorem

$(x-a)$ is said to be factor of $p(x)$ if and only if $p(a)=0$, where $p(x)$ is a polynomial of degree$n\ge 1$ and a is any real number.

Find the remainder when $\text{p(y)}=\text{12}{{\text{y}}^{\text{3}}}-\text{13}{{\text{y}}^{\text{2}}}-$$\text{6y}+\text{7}$ is divided by $\text{3y}+\text{4}$.

(a) $36\frac{5}{9}$

(b) $-36\frac{5}{9}$

(c) 36

(d) $-\frac{829}{9}$

(e) None of these

Explanation:

$3y+4=0$          $\Rightarrow$               $y=-\frac{4}{3}$

By remainder theorem, we know that p(y) is divided by 3y + 4 remainder will be $p\left( -\frac{4}{3} \right)$

$p\left( -\frac{4}{3} \right)=12\times {{\left( -\frac{4}{3} \right)}^{3}}-13{{\left( -\frac{4}{3} \right)}^{2}}-6\times {{\left( -\frac{4}{3} \right)}^{2}}+7$

$=\bcancel{{{12}^{4}}}\times \frac{-64}{\bcancel{{{27}_{9}}}}-13\times \frac{16}{9}+\frac{\bcancel{{{24}^{8}}}}{\bcancel{3}}+7=\frac{-256}{9}-\frac{208}{9}+15=\frac{-256-208+135}{9}=-\frac{329}{9}=-36\frac{5}{9}$

Is $(2y+1)$ a factor of $4{{y}^{3}}+4{{y}^{2}}-y-12$?

Solution:

$2y+1=0$          $\Rightarrow$               $y=-\frac{1}{2}$

Now, $f\left( -\frac{1}{2} \right)=4{{\left( -\frac{1}{2} \right)}^{3}}+4{{\left( -\frac{1}{2} \right)}^{2}}-\left( -\frac{1}{2} \right)-1$

$=\bcancel{4}\times \left( \frac{-1}{\bcancel{{{8}_{2}}}} \right)+4\times \frac{1}{{{\bcancel{4}}_{1}}}+\frac{1}{2}-1=\frac{-1}{2}+1+\frac{1}{2}-1=0$

Therefore, $(2y+1)$ is a factor of $4{{y}^{3}}+4{{y}^{2}}-y-1$

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