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ZEROES OF A POLYNOMIAL
A real number α is a zero of a polynomial p(x) if the value of the polynomial p(x) is zero at x = α. i.e. p(α) = 0
OR
The value of the variable x, for which the polynomial p(x) becomes zero is called zero of the polynomial.
Ex. : Consider, a polynomial p(x) = x^{2} – 5x + 6 ; replace x by 2 and 3.
p(2) = (2)^{2} – 5 × 2 + 6 = 4 – 10 + 6 = 0, p(3) = (3)^{2} – 5 × 3 + 6 = 9 – 15 + 6 = 0
∴ 2 and 3 are the zeroes of the polynomial p(x).
REMARK 1. The constant polynomial has no zero 2. Every linear polynomial has one and only one zero or root. 3. A given polynomial can have more than one zero or root. 4. If the degree of a polynomial is n, the maximum number of zeroes it can have is also n. Ex. : If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes, if the degree of polynomial is 8, maximum number of zeroes it can have is 8. 5. A zero of a polynomial need not be 0 6. 0 may be zero of the polynomial. 
Ex. Find which of the following algebraic expression is a polynomial.
(i) 3x^{2} – 5x
(iii) √y – 8
(iv) z^{5} – ∛z + 8
Sol.
(i) 3x^{2} – 5x = 3x^{2} – 5x^{1} Is a polynomial in one variable.
= x^{1} + x^{–1} Is not a polynomial as the second term 1/x has degree (–1).
(iii) √y – 8 = y^{1/2} – 8 Is not a polynomial since, the power of the first term (√y) is , which is not a whole number.
(iv) z^{5} – ∛z + 8 = z^{5} – z^{1/3} + 8
Since, the exponent of the second term is 1/3, which in not a whole number. Therefore, the given expression
is not a polynomial.
Ex. Find the degree of the polynomial :
(i) 5x^{2} – 6x^{3} + 8x^{7} + 6x^{2 }
(ii) 2y^{12} + 3y^{10} – y^{15} + y + 3
(iii) x
(iv) 8
Sol. (i) Since the term with highest exponent (power) is 8x^{7} and its power is 7.
∴ The degree of given polynomial is 7.
(ii) The highest power of the variable is 15. ⇒ degree = 15
(iii) x = x^{1} ⇒ degree is 1.
(iv) 8 = 8x^{0} ⇒ degree is 0.
Ex. Write the coefficient of x^{2 }in each of the following :
(i) 2 + x^{2} + x
(ii) 5 – x^{2} + x3
(iv) √5x – 1
Sol. (i) In the polynomial 2 + x^{2} + x the coefficient of x^{2} is 1.
(ii) In the polynomial 5 – x^{2} + x^{3} the coefficient of x^{2} is 1.
(iii) In the polynomial x^{2} + x the coefficient of x^{2} is.
(iv) In the polynomial √5x – 1 the coefficient of x^{2} = 0.
[We may write √5x – 1 = ax^{2} + √5x – 1. So the coefficient of x^{2} is 0]
GEOMETRICAL MEANING OF THE ZEROS OF A POLYNOMIAL
Geometrically the zeros of a polynomials f(x) are the xcoordinates of the points where the graph y = f(x) intersects xaxis. To understand it, we will see the geometrical representations of linear and quadratic polynomials.
GEOMETRICAL REPRESENTATION OF THE ZERO OF A LINEAR POLYNOMIAL
Consider a linear polynomial, y = 2x – 5.
The following table lists the values of y corresponding to different values of x.
x  1  4 
y  3  3 
On plotting the points A(1, –3) and B(4, 3) and joining them, a straight line is obtained. From, graph we observe that the graph of y = 2x – 5 intersects the xaxis at (5/2,0) whose xcoordinate is 5/2. Also, zero of 2x5 is 5/2.
Therefore, we conclude that the linear polynomial ax + b has one and only one zero, which is the xcoordinate of the point where the graph of y = ax + b intersects the xaxis
COMPETITION WINDOW
RELATIONSHIP BETWEEN THE ZEROS AND COEFFICIENTS OF A POLYNOMIAL
For a linear polynomial ax + b, , we have,
For a quadratic polynomial
If α and β are the zeros of a quadratic polynomial f(x). Then polynomial f(x) is given by
f(x) = K{x^{2} – (α + β)x + αβ}
or f(x) = K{x^{2} – (sum of the zeros) x + product of the zeros}
where K is a constant.
Ex. Find the zeroes of the quadratic polynomial 6x^{2} – 13x + 6 and verify the relation between the zeroes and its coefficients.
Sol. We have,6x^{2} – 13x + 6 = 6x^{2} – 4x – 9x + 6 = 2x(3x – 2) – 3(3x – 2) = (3x – 2) (2x – 3)
So, the value of 6x^{2} – 13x + 6 is 0, when (3x – 2) = 0 or (2x – 3) = 0 i.e.,
Ex. Find the zeroes of the quadratic polynomial 4x^{2} – 9 and verify the relation between the zeroes and its coefficients.
Sol. We have,4x^{2} – 9 = (2x)^{2} – (3)^{3 }= (2x – 3) (2x + 3)
So, the value of 4x^{2} – 9 is 0, when 2x – 3 = 0 or 2x + 3 = 0
REMAINDER THEOREM
If p(x) is any polynomial of degree greater than or equal to 1 and let a be any real number. When p(x) is divided by x– a, then the remainder is equal to p(a).
By division algorithm, we know that Dividend = Divisor × quotient + Remainder
d(x) = p(x) q(x) + r(x)
REMARK 1. If a polynomial p(x) is divided by (x + a), the remainder is equal to the value of p(x) at x = – a i.e., p(–a) 2. If a polynomial p(x) is divided by (ax – b), the remainder is equal to the value of p(x) at 3. If a polynomial p(x) is divided by (b – ax), the remainder is equal to the value of p(x) at 
Ex. Find the remainder when 4x^{3} – 3x^{2} + 2x – 4 is divided by :
(a) x – 1
(b) x + 2
(c) x +
Sol. Let p(x) = 4x^{3} – 3x^{2} + 2x – 4
(a) When p(x) is divided by (x – 1), then by remainder theorem, the required remainder will be p(1)
p(1) = 4(1)^{3} – 3(1)^{2} + 2(1) – 4 = 4 × 1 – 3 × 1 + 2 × 1 – 4 = 4 – 3 + 2 – 4 = – 1
(b) When p(x) is divided by (x + 2) then by remainder theorem, the required remainder will be p(–2).
p(–2) = 4(–2)^{3} – 3(–2)^{2} + 2(–2) – 4 = 4 × (–8) – 3 × 4 – 4 – 4 = – 32 – 12 – 8 = – 52
(c) When p(x) is divided by, then by remainder theorem, the required remainder will be
Ex. Find the remainder when the polynomial f(x) = 2x^{4} – 6x^{3} + 2x^{2} – x + 2 is divided by x + 2.
Sol. when f(x) is divided by (x + 2), then by remainder theorem remainder will be f(–2).
Now, f(x) = 2x^{4} – 6x^{3} + 2x^{2} – x + 2
⇒ f(–2) = 2(–2)^{4} – 6(–2)^{3} + 2(–2)^{2} – (–2) + 2
⇒ f(–2) = 2 × 16 – 6 × (–8) + 2 × 4 + 2 + 2
⇒ f(–2) = 32 + 48 + 8 + 2 + 2 = 92
Ex. If the polynomials ax^{3} + 4x^{2} + 3x – 4 and x^{3} – 4x + a leave the same remainder when divided by (x – 3), find the value of a.
Sol. Let p(x) = ax^{3} + 4x^{2} + 3x – 4 and q(x) = x^{3} – 4x + a be the given polynomials. The remainders when p(x) and q(x) are divided by (x – 3) are p(3) and q(3) respectively.
By the given condition, we have p(3) = q(3)
⇒ a × (3)^{3} + 4 × (3)^{2} + 3 × 3 – 4 = (3)^{3} – 4 × 3 + a
⇒ 27a + 36 + 9 – 4 = 27 – 12 + a
⇒ 26a + 26 = 0
⇒ 26a = – 26 ⇒ a = – 1
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