Algebra
Category : 8th Class
Introduction: Algebra is that branch of Mathematics in which letters represent any value which we can assign according to our requirement letters are generally of two types: constants and variables (or literal numbers).
POLYNOMIAL
Polynomial in one variable
An algebraic expression of the form \[(-2,3)\] where \[(\frac{1}{2},1)\] are constant and x is a variable; is called a polynomial expression in x.
Degree of polynomial in one variable: The degree of a polynomial in one variable is the greatest power (or index or exponent) of the variable.
Types of polynomial
Constant polynomial: If the polynomial is \[(2,\frac{1}{2})\], then it is called a constant polynomial and its degree is 0 (Zero).
Example: \[5{{x}^{2}}-7x-6=0\], etc, are polynomials of degree 0.
Linear polynomials in one variable: A polynomial of degree 1 in one variable is called linear polynomial. It is in the form of \[\left( -\frac{3}{5},2 \right)\]
Quadratic polynomial: Apolynomial of degree two in one variable is quadratic polynomial.
It i s in the form of \[(1,1)\]
Quadratic polynomial can be factorize in two linear factors.
Cubic polynomial: Apolynomial of degree three is called a cubic polynomial.
Polynomial in two or more variables
An algebraic expression containing two or more variables with the powers of the variables as non-negative integers (or whole numbers), is called a polynomial in two or more variables.
Example: \[\left( 2,-\frac{5}{3} \right)\] is a polynomial in two variables x and y.
Degrees of its terms are
\[(0,0)\] i.e., 2,3,4,4.
\[4{{x}^{2}}-20x+25=0\] The degree of the polynomial = 4.
Factors
When an algebraic expression can be written as the product of two or more expressions then each of these expressions is called a factor.
Factorization: The process of writing a given algebraic expression as the product of two or more factors is called factorization.
Important Formulae
Remainder theorem
If a polynomial P(x) is divided (x-a), then remainder is P(a), where a is any real number.
Example: If \[a+b-5,2a+ab+1\] is divided by x+2, then remainder is P(-2)
Now, we will put
\[a-b+5,2a-2b+5\] to get
\[a+b+c=9\]
\[ab+bc+ca=26,\]
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\]
\[x:\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}=3\]
Remainder = - 5
Factor theorem
If (x + a) is a factor of polynomial P(x), then remainder = 0
\[1/2(a+b+c)\] \[a+b+c\]
Example: Show that (x – 3) is a factor of the polynomial \[3(a+b+c)\]
If (x -3) is a factor of polynomial \[2(abc)\]then remainder \[x=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\]
\[{{x}^{3}}+\frac{1}{{{x}^{3}}}\]
\[3x-2=\frac{8}{x}.\]
\[\left( -\frac{3}{4},2 \right)\]
As remainder \[(2,2)\]
\[\left( -\frac{4}{3},2 \right)\] \[\frac{1}{x}-\frac{3}{4}+\frac{1}{2+x}=0\] is a factor of polynomial \[\left( 2,-\frac{3}{4} \right)\].
Linear Equation one variable
An equation involving a variable in first degree is called a linear equation in one variable. It is the form
\[(5,4)\] \[(-8,0)\] is variable
\[\left( -\frac{4}{3},2 \right)\]
\[\frac{x}{x-1}+\frac{x-1}{x}=2\frac{1}{2}\] is the root or solution of this equation.
Rules for solving a linear equation:
RULE 1 Same quantity (number) can be added to both sides of an equation without changing the equality.
RULE 2 Same quantity can be subtracted from both sides of an equation without changing the equality.
RULE 3 Both sides of an equation may be multiplied by the same non-zero number without changing the equality.
RULE 4 Both sides of an equation may be divided by the same non-zero number without changing the equality.
RULE 5 (Transposition) Any term of an equation may be taken to the other side with the sign changed. This process is called transposition.
Linear Equation in two variable
Example:
\[P(x)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+.....+{{a}_{n}}{{x}^{n}}\] …(i)
\[{{a}_{0}},{{a}_{1}},{{a}_{2}},.....,{{a}_{n}}\] ….(ii)
Put this value of x in (i)
\[{{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+.....+{{a}_{n}}{{x}^{n}}\]
\[{{a}_{0}},{{a}_{1}},{{a}_{2}}......{{a}_{n}}\]
\[({{a}_{0}}\ne 0)\]
\[-9,\frac{15}{4},\frac{-13}{3}\] \[f(x)=ax+b\]
\[f(x)=a{{x}^{2}}+bx+c\ne 0\]
\[3xy+7x{{y}^{2}}-8x{{y}^{3}}+7{{y}^{2}}{{x}^{2}}\]
\[1+1,1+2,1+3,2+2\]
\[\therefore \] \[{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] \[{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
\[{{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)\] Solution of this system of equations is \[{{(a-b)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)\]These two linear equations represent two lines interesting at the point (2, 1)
Quadratic Equation
Example (i): \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\]is a quadratic equation in x.
Example (ii): \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)\] is a quadratic equation my.
Example (iii): \[{{\text{a}}^{\text{3}}}\text{+}{{\text{b}}^{\text{3}}}\text{+}{{\text{c}}^{\text{3}}}\text{=3abc if a+b+c=0}\] and \[P(x)={{x}^{3}}+{{x}^{2}}+2x+3\]are not quadratic equations
\[x=-2\] is called second degree term, bx is called First degree- term c is called constant term or absolute term. a is called coefficient of \[P(x)={{x}^{3}}+{{x}^{2}}+2x+3\] is called coefficient of x.
A quadratic equation \[P(x)={{x}^{3}}-3{{x}^{2}}+4x-12\],\[P(x)={{x}^{3}}-3{{x}^{2}}+4x-12\] is said to be an incomplete quadratic equation if either b = 0 or c = 0 (or both are zero).
Example (i): \[P(3)=0\]is a complete quadratic equation
Example(ii): \[P(3)=0\] \[P(3)={{3}^{3}}-{{3}^{2}}+4\times 3-12\]are incomplete quadratic equation
Roots
\[27-27+12-12=0\] if it satisfies the equation.
i.e., if we replace x by a, both the sides of the equation become equal i.e., \[P(3)=0\]
Example (i): 2 is a root of the equation \[\therefore \]
[ \[(x-3)\] \[\frac{4-x}{7-x}=\frac{2}{5}\] i.e. \[20-5x=14-2x\] i.e., \[3x=6\] which is true]
Example (ii): 3 is not a root of the equation \[\Rightarrow \]
[\[x=2\] \[10x+x+3=11x+3\] i.e., \[=x+x+3=2x+3\]
i.e., \[\frac{11x+3}{2x+3}=\frac{4}{1}\] which is wrong]
Solution of a Quadratic Equation
Therefore a quadratic equation has either no real root or two equal real roots or two distinct real roots.
[It is to be noted here that all the quadratic equations cannot be solved by the method of factors. We will consider only those quadratic equations which can be solved by using factors.]
Method to solve a quadratic equation by using factors
The following steps are generally used:
(i) Write the quadratic equation as ax2 + bx + c = 0
i.e., making right hand side zero and clearing all the fractions on left hand side.
(ii) Factorise the left hand side into the product of linear factors.
(iii) Use the concept that ab = 0 implies either a = 0 or b =0 (or bother =0and^=0).
Therefore, put each linear factor equal to zero and solve the resulting linear equations.
We get the required solution.
Example: (i) \[11x+3=8x+12\]
Given; \[3x=9\]
\[2x=4\] \[11x+3=8x+12\] \[x=2\] \[f(x)=ax+b\]
\[f(x)=a{{x}^{2}}+bx+c\ne 0\]\[3xy+7x{{y}^{2}}-8x{{y}^{3}}+7{{y}^{2}}{{x}^{2}}\] or \[1+1,1+2,1+3,2+2\]
\[\therefore \] \[{{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] or \[{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
Thus, the roots are - 6 and 6.
Example: (ii) \[{{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)\]
\[{{(a-b)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)\] \[{{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}-ab+{{b}^{2}})\]
\[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})\] \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\]
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)\] \[{{\text{a}}^{\text{3}}}\text{+}{{\text{b}}^{\text{3}}}\text{+}{{\text{c}}^{\text{3}}}\text{=3abc if a+b+c=0}\]
\[P(x)={{x}^{3}}+{{x}^{2}}+2x+3\] \[x+2=0\] or \[x=-2\]
\[P(x)={{x}^{3}}+{{x}^{2}}+2x+3\] \[P(-2)={{(-2)}^{3}}+{{(-2)}^{2}}+2(-2)+3\] or \[=-8+4-4+3=-12+7=-5\]
Thus, the roots are -3, and -2.
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