**Polynomials - Class 10 : Notes**

**(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) = 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 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).

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 ≠ 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.

**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****: **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 ≠ 0, then we know that (x – Î±) and (x – 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}).

**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, 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: **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 = –b/a ,

Î±Î² + Î²Î³ + Î³Î± = c/a and Î± Î² Î³ = – 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).