Polynomials-Notes

A polynomial is a mathematical expression made up of variables, coefficients, and constants combined using addition, subtraction, multiplication, and non-negative integer exponents. Each part of the expression is called a term \[ ax^2 + bx + c\] Lets say first term = \[ ax^2\] \(a\) is coefficient of \(x^2\)

Second Term =\[bx\] \(b\) is coefficient of \(x\)

and constant term = \(c\)

The numbers associated with every variable term are called coefficients. hence in the Polynomials $$ ax^2 + bx + c$$ \(a\) and \(b\) are the coefficient where as c is constant term.

Powers raised to every variable are called exponents. hence in the Polynomials $$ ax^2 + bx + c$$ Thus, the power raised to x is the exponent. In this example \(2 ~ \& ~ 1\) are powers of x are exponents.

A zero polynomial is a special constant polynomial where all variable coefficients are zero, and its value is always zero, regardless of the variable's value. It is expressed as \(p(x) = 0\)

A monomial is a basic algebraic expression composed of a single term, which can be a constant, a variable, or a product of numbers and variables with non-negative integer exponents, e.g.
$$ 5x^2;\;x;\;3 $$

An algebraic expression consisting of exactly two non-zero terms, e.g. $$3x^2 +x;\quad x+2;\quady^3-5 $$

A trinomial is an algebraic expression that consists of exactly three non-zero terms separated by addition or subtraction signs, e.g. $$5x^2 + 6x + 5\\\\3y + 2z - 5$$

The degree of a polynomial is the highest exponent of any variable in the expression's terms with non-zero coefficients. coefficients, e.g. $$x^5 + y^4 + z -3\\ \Rightarrow\text{ Polynomial with degree: 5}\\\\z^3 -z^2 + 4z -5 \\\Rightarrow\text{Polynomial with degree: 3}$$

Degree of a non-zero constant polynomial is zero.

A quadratic polynomial is an algebraic expression with a highest degree of two. General form of a quadratic polynomial is $$ ax^2 + bx + c, \\\\~where ~ a, ~b, ~and ~c ~\text{ are coefficients and }\\\\ a \text{ cannot be zero}$$

A cubic polynomial is an algebraic expression where the variable's highest power is 3. General expression of cubic polynomial is$$ ax^3 + bx^2 + cx + d, ~where~ a\ne 0$$

A polynomial in one variable \(x\) of degree \(n\)is an expression of the form $$a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$$ where$$a_0, ~ a_1,~ a_2,~ \ldots~ a_n $$ are constants and $$a_n\ne0$$ In particular, if \[a_0 = a_1 = a_2 = a_3 = \ldots a_n=0\] (all the constants are zero), we get the Zero Polynomial.

The degree of zero polynomial is not defined.

The zeroes of a polynomial are the values of the variable that make the polynomial's value equal to zero.
For a polynomial \(p(x)\), a zero is any number \(c\) such that \(p(c)=0\).
These zeroes are also called roots or solutions of the polynomial.
Non-zero constant polynomial has no zero;
By convention every real number is a zero of zero polynomial

• A zero of the polynomial need not be zero
• 0 may be the zero of a polynomial
• Every linear polynomial has one and only one zero
• A polynomial can have more than one zero

Polynomial factorisation is the process of expressing a polynomial as a product of simpler, irreducible factors. Polynomial factors are similar to finding the prime factors of an integer.

If \(p(x)\) is a polynomial of degree \(n \geq 1\) and \(a\) is a real number, then

1. \((x-a)\) is a factor of \(P(x)\), if \(p(a)=0\), and
2. \(p(a)=0\), if \((x-a)\) is a factor of \(p(x)\)

Proof:

By Remainder Theorem, \[P(x)=(x-a) q(x) + p(a)\]

1. If \(p(a)=0\), then \(P(x)=(x-a)q(x)\), which shows that \((x-a)\) is a factor of \(p(x)\)


2. Since \(x-a\) is a factor of \(p(x),\;p(x)=(x-a)g(x)\) for same polynomial \(g(x)\).
   In this case, \(p(a)=(a-a)g(a)=0\)

Examine whether \(x+2\) is a factor of \(x^3 + 3x^2 + 5x +6\)

Solution:

The Zero of \(x+2 ~ is ~-2\) $$ \begin{aligned} Let~p(x) &= x^3 + 3x^2 + 5x +6\\\\ p(-2) &= (-2)^3 + 3(-2)^2 + 5(-2) +6\\\\ &= -8 -12-10+6\\\\ &=0 \end{aligned} $$ So, by Factor Theorem, \(x+2\) is a factor of \(x^3 + 3x^2 + 5x +6\)

Find the value of \(k\) if \(x-1\) is a factor of \(4x^3 + 3x^2 - 4x +k\)

Solution:

The Zero of \(x-1 \text{ is } 1\)

Let \(p(x) = 4x^3 + 3x^2 - 4x +k\)

because, \(x-1\) is a factor of \(p(x)\)

Hence, \(p(1) =0;\) $$ \require{cancel}\begin{aligned} p(1) &= 4(1)^3 + 3(1)^2 - 4(1) +k=0\\\\ &= 4+3-4+k=0\\\\ &= \cancel{4}+3-\cancel{4}+k=0\\\\ \Rightarrow 3k&=0\\\\ \Rightarrow k&=-3 \end{aligned} $$

Factorise \(6x^2 + 17x +5\) by splitting the middle term, and by using Factor Theorem.

Solution:

Factorisation by splitting method
\[\scriptsize \text{Let }(x+a)(x+b)=6x^2 + 17x +5 \tag{i} \] Expanding \((x+a)(x+b)\), we get $$\scriptsize (x+a)(x+b) = x^2 + (a+b)x + ab\\\tag{ii} $$ by comparing constant term of Eqn(i) and (ii) and coefficient of \(x\)
\[ab=6\times5=30\\\\a+b=17\]
Let us find possible values of \(a \;\&\; b\) for which \(a\times b=30\text{ and }a+b=17\) $$ \begin{array}{l|l} a&b\\\hline 1&30\\\color{orange}{2}&\color{orange}{15}\quad\Rightarrow\quad a+b=17\\3&10\\5&6 \end{array} $$ therefore we can write
\(6x^2 + 17x +5\) = \(6x^2 + (a+b)x + 5\)
$$ \begin{aligned} 6x^2 + 17x+5 &= 6x^2 + (2+15)x+5\\\\ &=6x^2 + 2x + 15x+5\\\\ &=2x(3x+1) +5(3x+2)\\\\ &=(3x+1)(2x+5) \end{aligned} $$

Factorise \(6x^2 + 17x +5\) by splitting the middle term, and by using Factor Theorem.

Solution:

Factorisation by splitting method
Let \[\scriptsize (x+a)(x+b)=6x^2 + 17x +5 \tag{i} \] Expanding \((x+a)(x+b)\), we get $$\scriptsize (x+a)(x+b) = x^2 + (a+b)x + ab\\\tag{ii} $$ by comparing constant term of Eqn(i) and (ii) and coefficient of \(x\)
\[ab=6\times5=30\\\\a+b=17\]
Let us find possible values of \(a ~\mathrm{\&}~ b\) for which \(a\times b=30 \text{ and }a+b=17\) $$ \begin{array}{l|l} a&b\\\hline 1&30\\\color{orange}{2}&\color{orange}{15}\quad\Rightarrow\quad a+b=17\\3&10\\5&6 \end{array} $$ therefore we can write
\(6x^2 + 17x +5=6x^2 + (a+b)x + 5\)
$$ \begin{aligned} 6x^2 + 17x+5 &= 6x^2 + (2+15)x+5\\\\ &=6x^2 + 2x + 15x+5\\\\ &=2x(3x+1) +5(3x+2)\\\\ &=(3x+1)(2x+5) \end{aligned} $$

Factorise by Factor Theorem
\[6x^2 + 15x +5\\\\ =6\left(x^2 + \frac{15}{6}x + \frac{5}{6}\right)=6p(x)\\\\ \Rightarrow p(x) = \left(x^2 + \frac{15}{6}x + \frac{5}{6}\right)\\\\ \text{Let }a ~\&~ b\text{ are the roots of } p(x)\\\\ \Rightarrow p(x) = x^2+(a+b) + ab\\=x^2 +\frac{17}{6}x +\frac{5}{6} \] We will now find all possible values of \(a~\&~b~(ab=\frac{5}{6})\) $$ \begin{array}{l|l} a&b\\\hline 1&\frac{5}{6}\\ \frac{1}{2}&\frac{5}{3}\\ \frac{1}{3}&\frac{5}{2} \end{array} $$ Possible Values = \(\pm 1, \pm\frac{1}{2} , \pm\frac{1}{3}, \pm\frac{5}{3}, \pm\frac{5}{2}\) So we will check only for negative values, as positive values for x will never become zero (reason being no negative term in the polynomial );

\(p(-\frac{1}{2})\) $$\scriptsize \begin{aligned} p(-\frac{1}{2}) &=\left(-\frac{1}{2}\right)^2 + \frac{17}{6}\times\left( -\frac{1}{2}\right) + \frac{5}{6}\\\\ &=\frac{1}{4}-\frac{17}{12}+\frac{5}{6}\\\\ &=\frac{3-17+10}{12}\\\\ &=-\frac{4}{12}\ne0 \end{aligned} $$
\(p(-\frac{1}{3})\)$$\scriptsize \begin{aligned} p(-\frac{1}{3}) &=\left(-\frac{1}{3}\right)^2 + \frac{17}{6}\times\left( -\frac{1}{3}\right) + \frac{5}{6}\\\\ &=\frac{1}{9}-\frac{17}{18}+\frac{5}{6}\\\\ &=\frac{2-17+15}{18}\\\\ &=-\frac{0}{18}\\\\&=0 \end{aligned} $$
Thus \(\left(x+\dfrac{1}{3}\right)\) is one factor

\(p(-\frac{5}{3})\)$$\scriptsize \begin{aligned} p(-\frac{5}{3}) &=\left(-\frac{5}{3}\right)^2 + \frac{17}{6}\times\left( -\frac{5}{3}\right) + \frac{5}{6}\\\\ &=\frac{25}{9}-\frac{85}{18}+\frac{5}{6}\\\\ &=\frac{50-85+15}{18}\\\\ &=-\frac{20}{18}\\\\&\ne0 \end{aligned} $$

\(p(-\frac{5}{2})\)$$\scriptsize \begin{aligned} p(-\frac{5}{2}) &=\left(-\frac{5}{2}\right)^2 + \frac{17}{6}\times\left( -\frac{5}{2}\right) + \frac{5}{6}\\\\ &=\frac{25}{4}-\frac{85}{12}+\frac{5}{6}\\\\ &=\frac{75-85+10}{18}\\\\ &=-\frac{0}{12}\\\\&=0 \end{aligned} $$
Therefore second factor =\(\left(x+\dfrac{5}{2}\right)\)
therefore, \[\scriptsize \require{cancel} \begin{aligned} 6p(x) = x^2 +17x +5&= 6\left(x+\frac{1}{3}\right)\left(x+\frac{5}{2}\right)\\\\ &=6\left(\frac{3x+1}{3}\right)\left(\frac{2x+5}{2}\right)\\\\ &=\cancel{6}\left(\frac{3x+1}{\cancel3}\right)\left(\frac{2x+5}{\cancel2}\right)\\\\ &=(3x+1)(2x+5) \end{aligned} \]

1. \((x+y)^2 = x^2 + 2xy +y^2\)
2. \((x-y)^2 = x^2 - 2xy +y^2\)
3. \(x^2-y^2 = (x+y)(x-y)\)
4. \((x+a)(x+b)=x^2 + (a+b)x + ab\)
5. \((x+y+z)^2=x^2+y^2+z^2 + 2xy+2yz+2x\)
6. \((x+y)^3 = x^3+y^3 + 3xy(x+y)\)
7. \((x-y)^3 = x^3-y^3 - 3xy(x-y)\)
8. \(x^3+y^3+z^3 -3xyz = (x+y+z)(x^2+y^2+z^2)xy-yz-zx\)

Evaluate 105 x 106 without multiplying directly

Solution
\[ \begin{aligned} 105 \times 106 &= (100 + 5) \times (100 + 6) \end{aligned}\] \[\scriptsize\color{orange}{\text{Using identity: } (x+a)(x+b) = x^2 + (a+b)x + ab}\] \[\small\begin{aligned}105 \times 106 &= (100 + 5) \times (100 + 6)\\&= 100^2 + (5+6) \times 100 + 5 \times 6 \\ &= 10000 + 11 \times 100 + 30 \\ &= 10000 + 1100 + 30 \\ &= 11130 \end{aligned} \]

Factorise \(\dfrac{25x^2}{4}-\dfrac{y^2}{9}\)

Solution
\[\color{orange}\text{Using Identity: }\\x^2-y^2=(x+y)(x-y)\] $$\scriptsize \begin{array}{l|r} \begin{aligned} &\qquad\frac{25}{4}x^2-\frac{y^2}{9}\\\\ &=\left(\frac{5}{2}x\right)^2-\left(\frac{y}{3}\right)^2&\\\\ &=\left(\frac{5x}{2}+\frac{y}{3}\right)\left(\frac{5x}{2}-\frac{y}{3}\right) \end{aligned} \end{array} $$

Evaluate \((109)^3\)

Solution
\( \begin{aligned} (109)^3&=(100+9)^3\\ &=100^3+9^3+3(100+9)100\times9\\\\ \end{aligned} \) \(\\\color{orange}\text{Using Identity: }\\(x+y)^3=x^3+y^3+3(x+y)xy\\ \begin{aligned}\color{violet} x&=100;\\y&=9\\\Rightarrow~x^3&=100^3\\&=1000000;\\y^3&=9^3\\&=729 \end{aligned}\) \[\small \begin{aligned} &=1000000+729+3\times(10+9)\times100\times9\\ &=1000000+729+3\times 9\times (100+9)\times100\\ &=1000000+729+3\times 9\times100\times (100+9)\\ &=1000729+2700\times(100+9)\\ &=1000729+(2700\times100)+(2700\times9)\\ &=1000729+270000+24300\\ &=1295029 \end{aligned} \]

Evaluate \((91)^3\)

Solution
$$ \begin{aligned} (91)^3&=(100-9)^3\\ &=100^3-9^3-3(100-9)100\times9 \end{aligned} $$ \(\\\color{orange}\text{Using Identity: }\\(x-y)^3=x^3-y^3-3(x-y)xy\\ \begin{aligned} x&=100;\\y&=9\\\Rightarrow~x^3&=100^3\\&=1000000;~\\y^3&=9^3\\&=729\end{aligned} \) \[\small \begin{aligned} &=1000000-729-3\times(100-9)\times100\times9\\ &=1000000-729-3\times 9\times (100-9)\times100\\ &=1000000-729-3\times 9\times100\times (100-9)\\ &=99271-2700\times(100-9)\\ &=99271-(2700\times100)+(2700\times9)\\ &=99271-270000+24300\\ &=753571 \end{aligned} \]

Factorise \(8x^3+27y^3+36x^2y+54xy^2\)

Solution
\[ 8x^3+27y^3+36x^2y+54xy^2\\ \\\text{Writing}\\\\ 8x^3 \text{ as } (2x)^2 \text{ and }\\\\ 27y^3 \text{ as } (3y)^3\\\\ \text{we can re-arrange the expression as }\\\\ \scriptsize(2x)^3+(3y)^3+3\cdot4\cdot x^2y+3\cdot2\cdot9xy^2\\\\ \scriptsize\Rightarrow(2x)^3+(3y)^3+3\cdot(2x)^2\cdot(3y)+3\cdot(2x)\cdot(3y)^2\\\\ \quad\color{orange}\text{Using Identity: }\\\\\scriptsize x^3+y^3+3x^2y+3xy^2=(x+y)^3\\\\ \scriptsize\Rightarrow (2x+3y)^3=(2x+3y)(2x+3y)(2x+3y) \]

1. A polynomial of one term is called a monomial.
2. A polynomial of two terms is called a binomial.
3. A polynomial of three terms is called a trinomial.
4. A polynomial of degree one is called a linear polynomial.
5. A polynomial of degree two is called a quadratic polynomial.
6. A polynomial of degree three is called a cubic polynomial.
7. A real number \(a\) is a zero of a polynomial \(p(x)~if~p(a) = 0\). In this case, a is also called a root of the equation \(p(x) = 0\).
8. Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial has no zero, and every real number is a zero of the zero polynomial.
9. Factor Theorem : \(x – a\) is a factor of the polynomial \(p(x), ~if ~p(a) = 0.\) Also, if \( x – a\) is a factor of \(p(x)\), then \(p(a) = 0\).