Types of Polynomials
Category : 9th Class
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
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\[{{(x-2)}^{2}}-{{(x-3)}^{2}}=2x-2\]]
Quadratic Polynomial
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
Biquadratic 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
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
Answer: (b)
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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