Notes for polynomials chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.
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(1) Polynomial :Â The expression which contains one or more terms with non-zero coefficient is called a polynomial. A polynomial can have any number of terms.
For Example:Â 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:Â The degree of p(x) = x^{5} â€“ x^{3} + 7 is 5.
(3) Linear polynomial : A polynomial of degree one is called a linear polynomial.
For Example:Â 1/(2x â€“ 7), âˆšs + 5, etc. are some linear polynomial.
(4) Quadratic polynomial : A polynomial having highest degree of two is called a quadratic polynomial. The term â€˜quadraticâ€™ is derived from word â€˜quadrateâ€™ which means square. In general, a quadratic polynomial can be expressed in the form ax^{2} + bx + c, where aâ‰ 0 and a, b, c are constants.
For Example:Â x^{2 }â€“ 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 aâ‰ 0 and a, b, c, d are constants.
For Example:Â x^{3 }â€“ 9x +2, a^{3} + a^{2} + âˆša + 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 â€“b/a = -(Constant term)/(coefficient of x)
For Example:Â 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).
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(7) Geometrical Meaning of the Zeroes of a Polynomial:
(i) For Linear Polynomial:
In general, for a linear polynomial ax + b, a â‰ 0, the graph of y = ax + b is a straight line which intersects the x-axis at exactly one point, namely, (-b/a , 0) . Therefore, the linear polynomial ax + b, a â‰ 0, has exactly one zero, namely, the x-coordinate of the point where the graph of y = ax + b intersects the x-axis.
For Example:Â 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 x-axis at exactly one point, namely, (3/2 , 0).
(ii) For Quadratic Polynomial:
In general, for any quadratic polynomial ax^{2} + bx + c, a â‰ 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 x-axis at two distinct points.The x-coordinates of the quadratic polynomial ax^{2} + bx + c have two zeros in this case.
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Case 2: The Graph cuts x-axis at exactly one point.The x-coordinates of the quadratic polynomial ax^{2} + bx + c have only one zero in this case.
Case 3: The Graph is completely above x-axis or below x-axis.The quadratic polynomial ax^{2} + bx + c have no zero in this case.
For Example:Â For the given graph, find the number of zeroes of p(x).From the figure, we can see that the graph intersects the x-axis 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 Î± and Î² are the zeroes of the quadratic polynomial p(x) = ax^{2} + bx + c, a â‰ 0, then we know that (x â€“ Î±) and (x â€“ Î²) are the factors of p(x).
Moreover, Î± + Î² = -b/a and Î± Î² = c/a.
In general, sum of zeros = -(Coefficient of x)/(Coefficient of x^{2}).
Product of zeros = (Constant term)/ (Coefficient of x^{2}).
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For Example:Â 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 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (-2) x (-5) = 10 = 10/1 = (Constant term)/ (Coefficient of x^{2}).
For Example:Â 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 + âˆš15) (t - âˆš15)
Thus, value of t^{2} -15 is zero for (t +âˆš15) = 0 or (t - âˆš15) = 0. Or in other words, for t = âˆš15 or t = -âˆš15.
Hence, zeros of t^{2} -15 are âˆš15 and -âˆš15.
Now, sum of zeros = âˆš15 + (-âˆš15) = 0 = -0/1 = -(Coefficient of t)/(Coefficient of t^{2}). Similarly, product of zeros = (âˆš15) x (-âˆš15) = -15 = -15/1 = (Constant term)/ (Coefficient of t^{2}).
For Example:Â 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, Î± + Î² = 4 = 4/1 = -b/a.
Î± Î² = 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:Â In general, it can be proved that if Î±, Î², Î³ are the zeroes of the cubic polynomial ax^{3} + bx^{2} + cx + d, then,
Î± + Î² + Î³ = â€“b/a ,
Î±Î² + Î²Î³ + Î³Î± = c/a andÂ Î± Î² Î³ = â€“ d/a .
(9) Division Algorithm for Polynomials : If p(x) and g(x) are any two polynomials with g(x) â‰ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) Ã— q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x).
For Example:Â Divide 3x^{2} â€“ x^{3} â€“ 3x + 5 by x â€“ 1 â€“ x^{2}, and verify the division algorithm.
On dividing 3x^{2} â€“ x^{3} â€“ 3x + 5 by x â€“ 1 â€“ x^{2}, we get,
Here, quotient is (x â€“ 2) and remainder is 3.
Now, as per the division algorithm, Divisor x Quotient + Remainder = Dividend
LHS = (-x^{2} + x + 1)(x â€“ 2) + 3
= (â€“x^{3} + x^{2} â€“ x + 2x^{2} â€“ 2x + 2 + 3)
= (â€“x^{3} + 3x^{2} â€“ 3x + 5)
RHS = (â€“x^{3} + 3x^{2} â€“ 3x + 5)
Thus, division algorithm is verified.
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For Example:Â On dividing x^{3} â€“ 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were (x â€“ 2) and (â€“2x + 4), respectively. Find g(x).
Given, dividend = p(x) = (x^{3} â€“ 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} â€“ 3x^{2} + x + 2) = g(x) (x â€“ 2) + (-2x + 4)
(x^{3} â€“ 3x^{2} + x + 2 + 2x -4) = g(x) (x â€“ 2)
(x^{3} â€“ 3x^{2} + 3x - 2) = g(x) (x â€“ 2)
Hence, g(x) is the quotient when we divide (x^{3} â€“ 3x^{2} + 3x - 2) by (x â€“ 2).Therefore, g(x) = (x^{2} â€“ x + 1).
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Okk
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