Relationship Between Zeros And Coefficients Of A Polynomial

Relationship Between Zeros And Coefficients Of A Polynomial

Consider quadratic polynomial
P(x) = 2x2 – 16x + 30.
Now, 2x2 – 16x + 30 = (2x – 6) (x – 3)
= 2 (x – 3) (x – 5)
The zeros of P(x) are 3 and 5.
Sum of the zeros = 3 + 5 = 8 = \(\frac { -\left( -16 \right)  }{ 2 }  \) = \(\text{-}\left[ \frac{\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)

Product of the zeros = 3 × 5 = 15 = \(\frac { 30 }{ 2 }\) = \(\left[ \frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)
So if ax2 + bx + c, a ≠ 0 is a quadratic polynomial and α, β are two zeros of polynomial then
\(\alpha +\beta =-\frac { b }{ a } \)
\(\alpha \beta =\frac { c }{ a } \)
In general, it can be proved that if α, β, γ are the zeros of a cubic polynomial ax3 + bx2 + cx + d, then
\(\alpha +\beta +\gamma =\frac { -b }{ a } \)
\( \alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \)
\( \alpha \beta \gamma =\frac { -d }{ a }  \)
Note:  \(\frac { b }{ a }  \), \(\frac { c }{ a }\) and \(\frac { d }{ a }  \)  are meaningful because a ≠ 0.

Relationship Between Zeros And Coefficients Of A Polynomial Example Problems With Solutions

Example 1:    Find the zeros of the quadratic polynomial 6x2 – 13x + 6 and verify the relation between the zeros and its coefficients.
Sol.    We have, 6x2 – 13x + 6 = 6×2 – 4x – 9x + 6
= 2x (3x – 2) –3 (3x – 2)
= (3x – 2) (2x – 3)
So, the value of 6x2 – 13x + 6 is 0, when
(3x – 2) = 0 or (2x – 3) = 0 i.e.,
When    x =  \(\frac { 2 }{ 3 }  \)  or   \(\frac { 3 }{ 2 }  \)
Therefore, the zeros of 6x2 – 13x + 6 are
\(\frac { 2 }{ 3 }  \)  and   \(\frac { 3 }{ 2 }  \)
Sum of the zeros
= \(\frac { 2 }{ 3 }  \) + \(\frac { 3 }{ 2 }  \) = \(\frac { 13 }{ 6 }  \) = \(\frac { \left( -13 \right)  }{ 6 } \) =  \(\text{-}\left[ \frac{\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)
Product of the zeros
= \(\frac { 2 }{ 3 }  \) × \(\frac { 3 }{ 2 }  \) = \(\frac { 6 }{ 6 }  \) = \(\left[ \frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)

Example 2:    Find the zeros of the quadratic polynomial 4x² – 9 and verify the relation between the zeros and its coefficients.
Sol.      We have,
4x2 – 9 = (2x)2 – 32 = (2x – 3) (2x + 3)
So, the value of 4x2 – 9 is 0, when
2x – 3 = 0   or   2x + 3 = 0
i.e., when   x = \(\frac { 3 }{ 2 }  \)   or   x = \(\frac { -3 }{ 2 }  \).
Therefore, the zeros of 4x2 – 9 are \(\frac { 3 }{ 2 }  \)   &  \(\frac { -3 }{ 2 }  \).
Sum of the zeros
=  \(\frac { 3 }{ 2 }\) \(-\frac { 3 }{ 2 }  \)  = 0 = \(-\frac { \left( 0 \right)  }{ 4 } \) = \(\text{-}\left[ \frac{\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)
Product of the zeros
= \(\frac { 3 }{ 2 }\) × \(\frac { -3 }{ 2 }\) = \(\frac { -9 }{ 4 }\) = \(\left[ \frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)

Example 3:    Find the zeros of the quadratic polynomial 9x2 – 5 and verify the relation between the zeros and its coefficients.
Sol.    We have,
9x2 – 5 = (3x)2 – (√5)2 = (3x – √5) (3x + √5)
So, the value of 9x2 – 5 is 0,
when 3x – √5 = 0 or 3x + √5 = 0
i.e., when x =  \(\frac { \sqrt { 5 }  }{ 3 } \)   or   x = \(\frac { -\sqrt { 5 }  }{ 3 } \).
Sum of the zeros
= \(\frac { \sqrt { 5 }  }{ 3 } \) \(-\frac { \sqrt { 5 }  }{ 3 } \) = 0 = \(-\frac { \left( 0 \right)  }{ 9 } \) = \(\text{-}\left[ \frac{\text{coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)
Product of the zeros
= \(\left( \frac { \sqrt { 5 }  }{ 3 }  \right)  \) × \(\left( \frac { -\sqrt { 5 }  }{ 3 }  \right)  \) =  \(\frac { -5 }{ 9 } \)  = \(\left[ \frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}} \right]\)

Example 4:    If α and β are the zeros of ax2 + bx + c, a ≠ 0 then verify the relation between the zeros and its coefficients.
Sol.    Since a and b are the zeros of polynomial ax2 + bx + c.
Therefore,   (x – α), (x – β) are the factors of the polynomial ax2 + bx + c.
⇒ ax2 + bx + c = k (x – α) (x – β)
⇒ ax2 + bx + c = k {x2 – (α + β) x + αβ}
⇒ ax2 + bx + c = kx2 – k (α + β) x + kαβ …(1)
Comparing the coefficients of x2, x and constant terms of (1) on both sides, we get
a = k, b = – k (α + β) and c = kαβ
⇒ α + β = \(\frac { -b }{ k }  \)      and        αβ = \(\frac { c }{ k }  \)
α + β = \(\frac { -b }{ a }  \)      and        αβ = \(\frac { c }{ a }  \) [∵  k = a]

Sum of the zeros = \(\frac { -b }{ a }  \) =  \(\frac{\text{- coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\)
Product of the zeros = \(\frac { c }{ a }  \) = \(\frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\)

Example 5:    Prove relation between the zeros and the coefficient of the quadratic polynomial
ax2 + bx + c.
Sol.    Let a and b be the zeros of the polynomial ax2 + bx + c
α = \(\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\)                   ….(1)
β = \(\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\)                   ….(2)
By adding (1) and (2), we get
α + β =  \(\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\) + \(\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\)
= \(\frac{ -2b }{ 2a } \) = \(\frac{ -b }{ a } \) = \(\frac{\text{- coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\)
Hence, sum of the zeros of the polynomial
ax2 + bx + c is  \(\frac{ -b }{ a } \)
By multiplying (1) and (2), we get
αβ =  \(\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\) × \(\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\)
= \(\frac{{{b}^{2}}-{{b}^{2}}+4ac}{4{{a}^{2}}}\)
= \(\frac{4ac}{4{{a}^{2}}}\) = \(\frac{ c }{ a } \)
= \(\frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\)
Hence, product of zeros = \(\frac{ c }{ a } \)

Example 6:    find the zeroes of the quadratic polynomial x2 – 2x – 8 and verify a relationship between zeroes and its coefficients.
Sol.    x2 – 2x – 8 = x2 – 4x + 2x – 8
= x (x – 4) + 2 (x – 4) = (x – 4) (x + 2)
So, the value of x2 – 2x – 8 is zero when
x – 4 = 0 or x + 2 = 0 i.e., when x = 4 or x = – 2.
So, the zeroes of x2 – 2x – 8 are 4, – 2.
Sum of the zeroes
= 4 – 2 = 2 = \(-\frac { \left( -2 \right)  }{ 1 } \) =  \(\frac{\text{- coefficient of x}}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\)
Product of the zeroes
= 4 (–2) = –8 = \(\frac { -8 }{ 1 }  \) = \(\frac{\text{constant term }}{\text{coefficient of }{{\text{x}}^{\text{2}}}}\)

Example 7:    Verify that the numbers given along side of the cubic polynomials are their zeroes. Also verify the relationship between the zeroes and the coefficients. 2x3 + x2 – 5x + 2 ; , 1, – 2
Sol.   Here, the polynomial p(x) is 2x3 + x2 – 5x + 2
Value of the polynomial 2x3 + x2 – 5x + 2
when x = 1/2
= \(2{{\left( \frac{1}{2} \right)}^{3}}+{{\left( \frac{1}{2} \right)}^{2}}-5\left( \frac{1}{2} \right)+2\) = \(\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2\) = 0
So, 1/2 is a zero of p(x).
On putting x = 1 in the cubic polynomial
2x3 + x2 – 5x + 2
= 2(1)3 + (1)2 –¬ 5(1) + 2 = 2 + 1 – 5 + 2 = 0
On putting x = – 2 in the cubic polynomial
2x3 + x2 – 5x + 2
= 2(–2)3 + (–2)2 – 5 (–2) + 2
= – 16 + 4 + 10 + 2 = 0
Hence, \(\frac{ 1 }{ 2 } \), 1, – 2 are the zeroes of the given polynomial.
Sum of the zeroes of p(x)
= \(\frac{ 1 }{ 2 } \) + 1 – 2 = \(\frac{ -1 }{ 2 } \) = \(\frac{-\text{ coefficient of }{{x}^{2}}}{\text{coefficient of }{{x}^{3}}}\)
Sum of the products of two zeroes taken at a time
= \(\frac{ 1 }{ 2 } \) × 1 + \(\frac{ 1 }{ 2 } \) × (–2) + 1 × (–2)
= \(\frac{ 1 }{ 2 } \) – 1 – 2 = \(\frac{ -5 }{ 2 } \) = \(\frac{\text{coefficient of }x}{\text{coefficient of }{{x}^{3}}}\)
Product of all the three zeroes
= \(\frac{ 1 }{ 2 } \) × (1) × (–2) = –1
= \(\frac{ -2 }{ 2 } \) = \(\frac{-\text{ constant term }}{\text{coefficient of }{{x}^{3}}}\)

Factorization Of Polynomials Using Factor Theorem

Factorization Of Polynomials Using Factor Theorem

Factor Theorem:

If p(x) is a polynomial of degree n  1 and a is any real number, then (i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x).
Proof: By the Remainder Theorem,
p(x) = (x – a) q(x) + p(a).
(i) If p(a) = 0, then p(x) = (x – a) q(x), which shows that x – a is a factor of p(x).
(ii) Since x – a is a factor of p(x),
p(x) = (x – a) g(x) for same polynomial g(x). In this case, p(a) = (a – a) g(a) = 0.

  1. Obtain the polynomial p(x).
  2. Obtain the constant term in p(x) and find its all possible factors. For example, in the polynomial
    x4 + x3 – 7x2 – x + 6 the constant term is 6 and its factors are ± 1, ± 2, ± 3, ± 6.
  3. Take one of the factors, say a and replace x by it in the given polynomial. If the polynomial reduces to zero, then (x – a) is a factor of polynomial.
  4. Obtain the factors equal in no. to the degree of polynomial. Let these are (x–a), (x–b), (x–c.)…..
  5. Write p(x) = k (x–a) (x–b) (x–c) ….. where k is constant.
  6. Substitute any value of x other than a,b,c …… and find the value of k.

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Factorization Of Polynomials Using Factor Theorem Example Problems With Solutions

Example 1:    Factorize x2 +4 + 9 z2 + 4x – 6 xz – 12 z
Solution:
The presence of the three squares viz.x2, (2)2, and (3z)2 gives a clue that identity (vii) could be used. So we write.
A = x2 + (2)2 + (3z)2 + 4x – 6 xz – 12 z
We note that the last two of the product terms are negative and that both of these contain z. Hence we write A as
A = x2 + (2)2 + (–3z)2 + 2.2x – 2.x.(–3z) + 2.2 (– 3z)
= (x+2 – 3z)2
= (x + 2 – 3z) (x + 2 – 3z)

Example 2:    Using factor theorem, factorize the polynomial x3 – 6x2 + 11 x – 6.
Solution:
Let f(x) = x3 – 6x2 + 11x – 6
The constant term in f(x) is equal to – 6 and factors of – 6 are ±1, ± 2, ± 3, ± 6.
Putting x = 1 in f(x), we have
f(1) = 13 – 6 ×12 + 11× 1– 6
= 1 – 6 + 11– 6 = 0
∴ (x– 1) is a factor of f(x)
Similarly, x – 2 and x – 3 are factors of f(x).
Since f(x) is a polynomial of degree 3. So, it can not have more than three linear factors.
Let f(x) = k (x–1) (x– 2) (x – 3). Then,
x3– 6x2 + 11x – 6 = k(x–1) (x– 2) (x– 3)
Putting x = 0 on both sides, we get
– 6 = k (0 – 1) (0 – 2) (0 – 3)
⇒ – 6 = – 6 k ⇒ k = 1
Putting k = 1 in f(x) = k (x– 1) (x– 2) (x–3), we get
f(x) = (x–1) (x– 2) (x – 3)
Hence, x3–6x2 + 11x – 6 = (x– 1) (x – 2) (x–3)

Example 3:    Using factor theorem, factorize the polynomial x4 + x3 – 7x2 – x + 6.
Solution:
Let f(x) = x4 + x3– 7x2 –x + 6
the factors of constant term in f(x) are ±1, ±2, ±3 and ± 6
Now,
Factorization Of Polynomials Using Factor Theorem 1
Since f(x) is a polynomial of degree 4. So, it cannot have more than 4 linear factors
Thus, the factors of f (x) are (x–1), (x+1),
(x–2) and (x+3).
Let f(x) = k (x–1) (x+1) (x–2) (x + 3)
⇒ x4 + x3 – 7x2 – x + 6
= k (x–1) (x +1) (x – 2) (x + 3)
Putting x = 0 on both sides, we get
6 = k (–1) (1) (–2) (3) ⇒ 6 = 6 k ⇒ k = 1
Substituting k = 1 in (i), we get
x4 + x3 – 7x2 – x + 6 = (x–1) (x +1) (x–2) (x+3)

Example 4:    Factorize,  2x4 + x3 – 14x2 – 19x – 6
Solution:
Let f(x) = 2x4 + x3 – 14x2 – 19x – 6 be the given polynomial. The factors of the constant term – 6 are ±1, ±2, ±3 and ±6, we have,
f(–1) = 2(–1)4 + (–1)3 – 14(–1)2 – 19(–1)– 6
= 2 – 1 – 14 + 19 – 6 = 21 – 21 = 0
and,
f(–2) = 2(–2)4 + (–2)3 – 14(–2)2 – 19(–2)– 6
= 32 – 8 – 56 + 38 – 6 = 0
So, x + 1 and x + 2 are factors of f(x).
⇒ (x + 1) (x + 2) is also a factor of f(x)
⇒ x2 + 3x + 2 is a factor of f(x)
Now, we divide
f(x) = 2x4 +x3 – 14x2–19x – 6 by
x2 + 3x + 2 to get the other factors.
Factorization Of Polynomials Using Factor Theorem 2

Example 5:    Factorize,  9z3 – 27z2 – 100 z+ 300, if it is given that (3z+10) is a factor of it.
Solution:
Let us divide 9z3 – 27z2 – 100 z+ 300 by
3z + 10 to get the other factors
Factorization Of Polynomials Using Factor Theorem 3
∴ 9z3 – 27z2 – 100 z+ 300
= (3z + 10) (3z2–19z + 30)
= (3z + 10) (3z2–10z – 9z + 30)
= (3z + 10) {(3z2–10z) – (9z – 30)}
= (3z + 10) {z(3z–10) – 3(3z–10)}
= (3z + 10) (3z–10) (z–3)
Hence, 9z3–27z2–100z+ 300
= (3z + 10) (3z–10) (z–3)

Example 6:    Simplify
\(\frac{4x-2}{{{x}^{2}}-x-2}+\frac{3}{2{{x}^{2}}-7x+6}-\frac{8x+3}{2{{x}^{2}}-x-3}\)
Solution:
Factorization Of Polynomials Using Factor Theorem 4

Example 7:    Establish the identity
\(\frac{6{{x}^{2}}+11x-8}{3x-2}=\left( 2x+5 \right)+\frac{2}{3x-2}\)
Solution:
Factorization Of Polynomials Using Factor Theorem 5

How To Form A Polynomial With The Given Zeroes

Form A Polynomial With The Given Zeros

Let zeros of a quadratic polynomial be α and β.
x = β,               x = β
x – α = 0,   x ­– β = 0
The obviously the quadratic polynomial is
(x – α) (x – β)
i.e.,  x2 – (α + β) x + αβ
x2 – (Sum of the zeros)x + Product of the zeros

Form A Polynomial With The Given Zeros Example Problems With Solutions

Example 1:    Form the quadratic polynomial whose zeros are 4 and 6.
Sol.    Sum of the zeros = 4 + 6 = 10
Product of the zeros = 4 × 6 = 24
Hence the polynomial formed
= x2 – (sum of zeros) x + Product of zeros
= x2 – 10x + 24

Example 2:    Form the quadratic polynomial whose zeros are –3, 5.
Sol.    Here, zeros are – 3 and 5.
Sum of the zeros = – 3 + 5 = 2
Product of the zeros = (–3) × 5 = – 15
Hence the polynomial formed
= x2 – (sum of zeros) x + Product of zeros
= x2 – 2x – 15

Example 3:    Find a quadratic polynomial whose sum of zeros and product of zeros are respectively \(\frac { 1 }{ 2 }\), – 1
Sol.   Let the polynomial be ax2 + bx + c and its zeros be  α and β.
(i) Here, α + β = \(\frac { 1 }{ 4 }\) and α.β = – 1
Thus the polynomial formed
= x2 – (Sum of zeros) x + Product of zeros
\(={{\text{x}}^{\text{2}}}-\left( \frac{1}{4} \right)\text{x}-1={{\text{x}}^{\text{2}}}-\frac{\text{x}}{\text{4}}-1\)
The other polynomial are   \(\text{k}\left( {{\text{x}}^{\text{2}}}\text{-}\frac{\text{x}}{\text{4}}\text{-1} \right)\)
If k = 4, then the polynomial is 4x2 – x – 4.

Example 4:    Find a quadratic polynomial whose sum of zeros and product of zeros are respectively \(\sqrt { 2 }\),  \(\frac { 1 }{ 3 }\)
Sol. Here, α + β =\(\sqrt { 2 }\), αβ = \(\frac { 1 }{ 3 }\)
Thus the polynomial formed
= x2 – (Sum of zeroes) x + Product of zeroes
= x2 – \(\sqrt { 2 }\) x + \(\frac { 1 }{ 3 }\)
Other polynomial are   \(\text{k}\left( {{\text{x}}^{\text{2}}}\text{-}\frac{\text{x}}{\text{3}}\text{-1} \right)\)
If k = 3, then the polynomial is
3x2 – \(3\sqrt { 2 }x\)  + 1

Example 5:    Find a quadratic polynomial whose sum of zeros and product of zeros are respectively 0, √5
Sol. Here, α + β = 0, αβ = √5
Thus the polynomial formed
= x2 – (Sum of zeroes) x + Product of zeroes
= x2 – (0) x + √5 = x2 + √5

Example 6:    Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken two at a time, and product of its zeroes as 2, – 7 and –14, respectively.
Sol.    Let the cubic polynomial be ax3 + bx2 + cx + d
⇒ x3 + \(\frac { b }{ a }\)x2 + \(\frac { c }{ a }\)x + \(\frac { d }{ a }\) …(1)
and its zeroes are α, β and γ then
α + β + γ = 2 = \(\frac { -b }{ a }\)
αβ + βγ + γα = – 7 = \(\frac { c }{ a }\)
αβγ = – 14 = \(\frac { -d }{ a }\)
Putting the values of   \(\frac { b }{ a }\), \(\frac { c }{ a }\),  and \(\frac { d }{ a }\)  in (1), we get
x3 + (–2) x2 + (–7)x + 14
⇒ x3 – 2x2 – 7x + 14

Example 7:   Find the cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and product of its zeroes as 0, –7 and –6 respectively.
Sol.    Let the cubic polynomial be ax3 + bx2 + cx + d
⇒ x3 + \(\frac { b }{ a }\)x2 + \(\frac { c }{ a }\)x + \(\frac { d }{ a }\) …(1)
and its zeroes are α, β and γ then
α + β + γ = 0 = \(\frac { -b }{ a }\)
αβ + βγ + γα = – 7 = \(\frac { c }{ a }\)
αβγ = – 6 = \(\frac { -d }{ a }\)
Putting the values of   \(\frac { b }{ a }\), \(\frac { c }{ a }\),  and \(\frac { d }{ a }\)  in (1), we get
x3 – (0) x2 + (–7)x + (–6)
⇒ x3 – 7x + 6

Example 8:   If α and β are the zeroes of the polynomials  ax2 + bx + c then form the polynomial whose zeroes are    \(\frac { 1 }{ \alpha  } \quad and\quad \frac { 1 }{ \beta  } \)
Since α and β are the zeroes of ax2 + bx + c
So α + β = \(\frac { -b }{ a }\) ,     α β =  \(\frac { c }{ a }\)
Sum of the zeroes = \(\frac { 1 }{ \alpha  } +\frac { 1 }{ \beta  } =\frac { \alpha +\beta  }{ \alpha \beta  }  \)
\(=\frac{\frac{-b}{c}}{\frac{c}{a}}=\frac{-b}{c}\)
Product of the zeroes
\(=\frac{1}{\alpha }.\frac{1}{\beta }=\frac{1}{\frac{c}{a}}=\frac{a}{c}\)
But required polynomial is
x2 – (sum of zeroes) x + Product of zeroes
\(\Rightarrow {{\text{x}}^{2}}-\left( \frac{-b}{c} \right)\text{x}+\left( \frac{a}{c} \right)\)
\(\Rightarrow {{\text{x}}^{2}}+\frac{b}{c}\text{x}+\frac{a}{c}\)
\(\Rightarrow c\left( {{\text{x}}^{2}}+\frac{b}{c}\text{x}+\frac{a}{c} \right)\)
⇒ cx2 + bx + a

How Do You Determine The Degree Of A Polynomial

Degree Of A Polynomial

The greatest power (exponent) of the terms of a polynomial is called degree of the polynomial.
For example :
In polynomial 5x2 – 8x7 + 3x:
(i) The power of term 5x2 = 2
(ii) The power of term –8x7 = 7
(iii) The power of 3x = 1
Since, the greatest power is 7, therefore degree of the polynomial 5x2 – 8x7 + 3x is 7
The degree of polynomial :
(i) 4y3 – 3y + 8 is 3
(ii) 7p + 2 is 1(p = p1)
(iii) 2m – 7m8 + m13 is 13 and so on.

Degree Of A Polynomial With Example Problems With Solutions

Example 1:    Find which of the following algebraic expression is a polynomial.
(i) 3x2 – 5x        (ii) \(\text{x + }\frac{1}{\text{x}}\)     (iii) √y– 8             (iv) z5 – ∛z + 8
Sol.
(i) 3x2 – 5x = 3x2 – 5x1
It is a polynomial.
(ii) \(\text{x + }\frac{1}{\text{x}}\) = x1 + x-1
It is not a polynomial.
(iii) √y– 8 = y1/2– 8
Since, the power of the first term (√y) is \(\frac{1}{2}\), which is not a whole number.
(iv) z5 – ∛z + 8 = z5 – z1/3 + 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.

Example 2:    Find the degree of the polynomial :
(i) 5x – 6x3 + 8x7 + 6x2    (ii) 2y12 + 3y10 – y15 + y + 3   (iii) x    (iv) 8
Sol.
(i)  Since the term with highest exponent (power) is 8x7 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 = x1   ⇒   degree is 1.
(iv)  8 = 8x0   ⇒   degree = 0

Zeros Of A Polynomial Function

Zeros Of A Polynomial Function

If for x = a, the value of the polynomial p(x) is 0 i.e., p(a) = 0; then x = a is a zero of the polynomial p(x).

For Example:
(i) For polynomial p(x) = x – 2; p(2) = 2 – 2 = 0
∴ x = 2 or simply 2 is a zero of the polynomial
p(x) = x – 2.
(ii) For the polynomial g(u) = u2 – 5u + 6;
g(3) = (3)2 – 5 × 3 + 6 = 9 – 15 + 6 = 0
∴ 3 is a zero of the polynomial g(u)
= u2 – 5u + 6.
Also, g(2) = (2)2 – 5 × 2 + 6 = 4 – 10 + 6 = 0
∴ 2 is also a zero of the polynomial
g(u) = u2 – 5u + 6
(a) Every linear polynomial has one and only one zero.
(b) A given polynomial may have more than one zeroes.
(c) If the degree of a polynomial is n; the largest number of zeroes it can have is also n.
For Example:
If the degree of a polynomial is 5, the polynomial can have at the most 5 zeroes; if the degree of a
polynomial is 8; largest number of zeroes it can have is 8.
(d) A zero of a polynomial need not be 0.
For Example: If f(x) = x2 – 4,
then f(2) = (2)2 – 4 = 4 – 4 = 0
Here, zero of the polynomial f(x) = x2 – 4 is 2 which itself is not 0.
(e) 0 may be a zero of a polynomial.
For Example: If f(x) = x2 – x,
then f(0) = 02 – 0 = 0
Here 0 is the zero of polynomial
f(x) = x2 – x.

Zeros Of A Polynomial Function With Examples

Example 1:    Verify whether the indicated numbers are zeroes of the polynomial corresponding to them in the following cases :
(i) p(x) = 3x + 1, x = \(-\frac{1}{3}\)
(ii) p(x) = (x + 1) (x – 2), x = – 1, 2
(iii) p(x) = x2, x = 0
(iv) p(x) = lx + m, x = \(-\frac{m}{l}\)
(v) p(x) = 2x + 1, x = \(\frac{1}{2}\)
Sol.
(i) p(x) = 3x + 1
\(\Rightarrow p\left( -\frac{1}{3} \right)=3\times -\frac{1}{3}+1=-1+1=0\)
∴ x = \(-\frac{1}{3}\)  is a zero of p(x) = 3x + 1.
(ii) p(x) = (x + 1) (x – 2)
⇒ p(–1) = (–1 + 1) (–1 – 2) = 0 × –3 = 0
and, p(2) = (2 + 1) (2 – 2) = 3 × 0 = 0
∴  x = –1 and x = 2 are zeroes of the given polynomial.
(iii) p(x) = x
⇒ p(0) = 02 = 0
∴  x = 0 is a zero of the given polynomial
(iv) p(x) = lx + m
\(\Rightarrow p\left( -\frac{m}{l} \right)=l\left( -\frac{m}{l} \right)+m\)
= – m + m = 0
∴  x = \(-\frac{m}{l}\)  is a zero of the given polynomial.
(v) p(x) = 2x + 1
\(\Rightarrow p\left( \frac{1}{2} \right)=2\times \frac{1}{2}+1\)
= 1 + 1 = 2 ≠ 0
∴ x = \(\frac{1}{2}\) is not a zero of the given polynomial.

Example 2:    Find the zero of the polynomial in each of the following cases :
(i) p(x) = x + 5
(ii) p(x) = 2x + 5
(iii) p(x) = 3x – 2
Sol.
To find the zero of a polynomial p(x) means to solve the polynomial equation p(x) = 0.
(i) For the zero of polynomial p(x) = x + 5
p(x) = 0      ⇒   x + 5 = 0     ⇒   x = –5
∴   x = –5 is a zero of the polynomial.
p(x) = x + 5.
(ii) p(x) = 0 ⇒ 2x + 5 = 0
⇒ 2x = –5 and x = \(-\frac{5}{2}\)
∴  \(-\frac{5}{2}\) is a zero of p(x) = 2x + 5.
(iii) p(x) = 0 ⇒ 3x – 2 = 0
⇒ 3x = 2 and x = \(\frac{2}{3}\).
∴  x = \(\frac{2}{3}\) is zero of p(x) = 3x – 2.