# 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{2}}}-left( frac{1}{4} right)text{x}-1=^{text{2}}}-frac{text{x}}{text{4}}-1)

The other polynomial are   (text{k}left( ^{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{2}}}text{-}frac{text{x}}{text{3}}text{-1} right))

If k = 3, then the polynomial is

3x2 – (3sqrt { 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 andquad 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 ^{2}}-left( frac{-b}{c} right)text{x}+left( frac{a}{c} right))

(Rightarrow ^{2}}+frac{b}{c}text{x}+frac{a}{c})

(Rightarrow cleft( ^{2}}+frac{b}{c}text{x}+frac{a}{c} right))

⇒ cx2 + bx + a

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