
01. Real Numbers
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Lecture1.4

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02. Polynomials
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03. Linear Equation
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04. Quadratic Equation
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05. Arithmetic Progressions
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06. Some Applications of Trigonometry
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07. Coordinate Geometry
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08. Triangles
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09. Circles
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10. Areas Related to Circles
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11. Introduction to Trigonometry
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12. Surface Areas and Volumes
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13. Statistics
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14. Probability
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15. Construction
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Lecture15.7

Chapter Notes – Polynomials
(1) Polynomial : The expression which contains one or more terms with nonzero coefficient is called a polynomial. A polynomial can have any number of terms.
For Example: 10, a + b, 7x + y + 5, w + x + y + z, etc. are some polynomials.
(2) Degree of polynomial : The highest power of the variable in a polynomial is called as the degree of the polynomial.
For Example: The degree of p(x) = x^{5} – x^{3} + 7 is 5.
(3) Linear polynomial : A polynomial of degree one is called a linear polynomial.
For Example: 1/(2x – 7), √s + 5, etc. are some linear polynomial.
(4) Quadratic polynomial : A polynomial having highest degree of two is called a quadratic polynomial. The term ‘quadratic’ is derived from word ‘quadrate’ which means square. In general, a quadratic polynomial can be expressed in the form ax^{2} + bx + c, where a≠0 and a, b, c are constants.
For Example: x^{2 }– 9, a^{2} + a + 7, etc. are some quadratic polynomials.
(5) Cubic Polynomial : A polynomial having highest degree of three is called a cubic polynomial. In general, a quadratic polynomial can be expressed in the form ax^{3} + bx^{2} + cx + d, where a≠0 and a, b, c, d are constants.
For Example: x^{3 }– 9x +2, a^{3} + a^{2} + √a + 7, etc. are some cubic polynomial.
(6) Zeroes of a Polynomial : The value of variable for which the polynomial becomes zero is called as the zeroes of the polynomial. In general, if k is a zero of p(x) = ax + b, then p(k) = ak + b = 0, i.e., k = b/a. Hence, the zero of the linear polynomial ax + b is –b/a = (Constant term)/(coefficient of x)
For Example: Consider p(x) = x + 2. Find zeroes of this polynomial.
If we put x = 2 in p(x), we get,
p(2) = 2 + 2 = 0.
Thus, 2 is a zero of the polynomial p(x).
(7) Geometrical Meaning of the Zeroes of a Polynomial:
(i) For Linear Polynomial:
In general, for a linear polynomial ax + b, a ≠ 0, the graph of y = ax + b is a straight line which intersects the xaxis at exactly one point, namely, (b/a , 0) . Therefore, the linear polynomial ax + b, a ≠ 0, has exactly one zero, namely, the xcoordinate of the point where the graph of y = ax + b intersects the xaxis.
For Example: The graph of y = 2x – 3 is a straight line passing through points (0, 3) and (3/2, 0).
x  0  3/2 
y = 2x – 3  6  0 
Here, the graph of y = 2x – 3 is a straight line which intersects the xaxis at exactly one point, namely, (3/2 , 0).
(ii) For Quadratic Polynomial:
In general, for any quadratic polynomial ax^{2} + bx + c, a ≠ 0, the graph of the corresponding equation y = ax^{2} + bx + c has one of the two shapes either open upwards like curve or open downwards like curve depending on whether a > 0 or a < 0. (These curves are called parabolas.)
Case 1: The Graph cuts xaxis at two distinct points.The xcoordinates of the quadratic polynomial ax^{2} + bx + c have two zeros in this case.
Case 2: The Graph cuts xaxis at exactly one point.The xcoordinates of the quadratic polynomial ax^{2} + bx + c have only one zero in this case.
Case 3: The Graph is completely above xaxis or below xaxis.The quadratic polynomial ax^{2} + bx + c have no zero in this case.
For Example: For the given graph, find the number of zeroes of p(x).From the figure, we can see that the graph intersects the xaxis at four points.
Therefore, the number of zeroes is 4.
(8) Relationship between Zeroes and Coefficients of a Polynomial:
(i) Quadratic Polynomial:
In general, if α and β are the zeroes of the quadratic polynomial p(x) = ax^{2} + bx + c, a ≠ 0, then we know that (x – α) and (x – β) are the factors of p(x).
Moreover, α + β = b/a and α β = c/a.
In general, sum of zeros = (Coefficient of x)/(Coefficient of x^{2}).
Product of zeros = (Constant term)/ (Coefficient of x^{2}).
For Example: Find the zeroes of the quadratic polynomial x^{2} + 7x + 10, and verify the relationship between the zeroes and the coefficients.
On finding the factors of x^{2} + 7x + 10, we get, x^{2}+ 7x + 10 = (x + 2) (x + 5)
Thus, value of x^{2} + 7x + 10 is zero for (x+2) = 0 or (x +5)= 0. Or in other words, for x = 2 or x = 5.
Hence, zeros of x^{2} + 7x + 10 are 2 and 5.
Now, sum of zeros = 2 + (5) = 7 = 7/1 = (Coefficient of x)/(Coefficient of x^{2}). Similarly, product of zeros = (2) x (5) = 10 = 10/1 = (Constant term)/ (Coefficient of x^{2}).
For Example: Find the zeroes of the quadratic polynomial t^{2} 15, and verify the relationship between the zeroes and the coefficients.
On finding the factors of t^{2} 15, we get, t^{2} 15= (t + √15) (t – √15)
Thus, value of t^{2} 15 is zero for (t +√15) = 0 or (t – √15) = 0. Or in other words, for t = √15 or t = √15.
Hence, zeros of t^{2} 15 are √15 and √15.
Now, sum of zeros = √15 + (√15) = 0 = 0/1 = (Coefficient of t)/(Coefficient of t^{2}). Similarly, product of zeros = (√15) x (√15) = 15 = 15/1 = (Constant term)/ (Coefficient of t^{2}).
For Example: Find a quadratic polynomial for the given numbers as the sum and product of its zeroes respectively 4, 1.
Let the quadratic polynomial be ax^{2} + bx + c.
Given, α + β = 4 = 4/1 = b/a.
α β = 1 = 1/1 = c/a.
Thus, a = 1, b = 4 and c = 1.
Therefore, the quadratic polynomial is x^{2} – 4x + 1.
(ii) Cubic Polynomial: In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial ax^{3} + bx^{2} + cx + d, then,
α + β + γ = –b/a ,
αβ + βγ + γα = c/a and α β γ = – d/a .
(9) Division Algorithm for Polynomials : If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x).
For Example: Divide 3x^{2} – x^{3} – 3x + 5 by x – 1 – x^{2}, and verify the division algorithm.
On dividing 3x^{2} – x^{3} – 3x + 5 by x – 1 – x^{2}, we get,
Here, quotient is (x – 2) and remainder is 3.
Now, as per the division algorithm, Divisor x Quotient + Remainder = Dividend
LHS = (x^{2} + x + 1)(x – 2) + 3
= (–x^{3} + x^{2} – x + 2x^{2} – 2x + 2 + 3)
= (–x^{3} + 3x^{2} – 3x + 5)
RHS = (–x^{3} + 3x^{2} – 3x + 5)
Thus, division algorithm is verified.
For Example: On dividing x^{3} – 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were (x – 2) and (–2x + 4), respectively. Find g(x).
Given, dividend = p(x) = (x^{3} – 3x^{2} + x + 2), quotient = (x 2), remainder = (2x + 4).
Let divisor be denoted by g(x).
Now, as per the division algorithm,
Divisor x Quotient + Remainder = Dividend
(x^{3} – 3x^{2} + x + 2) = g(x) (x – 2) + (2x + 4)
(x^{3} – 3x^{2} + x + 2 + 2x 4) = g(x) (x – 2)
(x^{3} – 3x^{2} + 3x – 2) = g(x) (x – 2)
Hence, g(x) is the quotient when we divide (x^{3} – 3x^{2} + 3x – 2) by (x – 2).Therefore, g(x) = (x^{2} – x + 1).
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