 # Relationship Between Zeros And Coefficients Of A Polynomial  ## Relationship Between Zeros And Coefficients Of A 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{2}}}} right])

Product of the zeros = 3 × 5 = 15 = (frac { 30 }{ 2 }) = (left[ frac{text{constant term }}{text{coefficient of }^{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{2}}}} right])

Product of the zeros

= (frac { 2 }{ 3 }  ) × (frac { 3 }{ 2 }  ) = (frac { 6 }{ 6 }  ) = (left[ frac{text{constant term }}{text{coefficient of }^{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{2}}}} right])

Product of the zeros

= (frac { 3 }{ 2 }) × (frac { -3 }{ 2 }) = (frac { -9 }{ 4 }) = (left[ frac{text{constant term }}{text{coefficient of }^{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{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{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{2}}}})

Product of the zeros = (frac { c }{ a }  ) = (frac{text{constant term }}{text{coefficient of }^{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-4ac}}{2a})                   ….(1)

β = (frac{-b-sqrt-4ac}}{2a})                   ….(2)

By adding (1) and (2), we get

α + β =  (frac{-b+sqrt-4ac}}{2a}) + (frac{-b-sqrt-4ac}}{2a})

= (frac{ -2b }{ 2a } ) = (frac{ -b }{ a } ) = (frac{text{- coefficient of x}}{text{coefficient of }^{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-4ac}}{2a}) × (frac{-b-sqrt-4ac}}{2a})

= (frac-+4ac}{4})

= (frac{4ac}{4}) = (frac{ c }{ a } )

= (frac{text{constant term }}{text{coefficient of }^{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{2}}}})

Product of the zeroes

= 4 (–2) = –8 = (frac { -8 }{ 1 }  ) = (frac{text{constant term }}{text{coefficient of }^{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+-5left( 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 }}{text{coefficient of }})

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 }})

Product of all the three zeroes

= (frac{ 1 }{ 2 } ) × (1) × (–2) = –1

= (frac{ -2 }{ 2 } ) = (frac{-text{ constant term }}{text{coefficient of }}) You might also like 