Chapter Notes – Polynomials

(1) Polynomial : 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: 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) = x⁵ – x³ + 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 ax² + bx + c, where a≠0 and a, b, c are constants.
For Example: x² – 9, a² + 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 ax³ + bx² + cx + d, where a≠0 and a, b, c, d are constants.
For Example: x³ – 9x +2, a³ + a² + √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 x-axis at exactly one point, namely, (-b/a , 0) . Therefore, the linear polynomial ax + b, a ≠ 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: The graph of y = 2x – 3 is a straight line passing through points (0, -3) and (3/2, 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 ax² + bx + c, a ≠ 0, the graph of the corresponding equation y = ax² + 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 ax² + bx + c have two zeros in this case.

Case 2: The Graph cuts x-axis at exactly one point.The x-coordinates of the quadratic polynomial ax² + 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 ax² + 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 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 α and β are the zeroes of the quadratic polynomial p(x) = ax² + 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 x²).
Product of zeros = (Constant term)/ (Coefficient of x²).

For Example: Find the zeroes of the quadratic polynomial x² + 7x + 10, and verify the relationship between the zeroes and the coefficients.
On finding the factors of x² + 7x + 10, we get, x²+ 7x + 10 = (x + 2) (x + 5)
Thus, value of x² + 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 x² + 7x + 10 are -2 and -5.
Now, sum of zeros = -2 + (-5) = -7 = -7/1 = -(Coefficient of x)/(Coefficient of x²). Similarly, product of zeros                                = (-2) x (-5) = 10 = 10/1 = (Constant term)/ (Coefficient of x²).

For Example: Find the zeroes of the quadratic polynomial t² -15, and verify the relationship between the zeroes and the coefficients.
On finding the factors of t² -15, we get, t² -15= (t + √15) (t – √15)
Thus, value of t² -15 is zero for (t +√15) = 0 or (t – √15) = 0. Or in other words, for t = √15 or t = -√15.
Hence, zeros of t² -15 are √15 and -√15.
Now, sum of zeros = √15 + (-√15) = 0 = -0/1 = -(Coefficient of t)/(Coefficient of t²). Similarly, product of zeros = (√15) x (-√15) = -15 = -15/1 = (Constant term)/ (Coefficient of t²).

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 ax² + 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 x² – 4x + 1.

(ii) Cubic Polynomial: In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + 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 3x² – x³ – 3x + 5 by x – 1 – x², and verify the division algorithm.
On dividing 3x² – x³ – 3x + 5 by x – 1 – x², we get,

Here, quotient is (x – 2) and remainder is 3.
Now, as per the division algorithm, Divisor x Quotient + Remainder = Dividend
LHS = (-x² + x + 1)(x – 2) + 3
= (–x³ + x² – x + 2x² – 2x + 2 + 3)
= (–x³ + 3x² – 3x + 5)
RHS = (–x³ + 3x² – 3x + 5)
Thus, division algorithm is verified.

For Example: On dividing x³ – 3x² + 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) = (x³ – 3x² + 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
(x³ – 3x² + x + 2) = g(x) (x – 2) + (-2x + 4)
(x³ – 3x² + x + 2 + 2x -4) = g(x) (x – 2)
(x³ – 3x² + 3x – 2) = g(x) (x – 2)
Hence, g(x) is the quotient when we divide (x³ – 3x² + 3x – 2) by (x – 2).Therefore, g(x) = (x² – x + 1).