**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., x^{2} – (α + β) x + αβ

x^{2} – (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

= x^{2} – (sum of zeros) x + Product of zeros

= x^{2} – 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

= x^{2} – (sum of zeros) x + Product of zeros

= x^{2} – 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 ax^{2} + bx + c and its zeros be α and β.

(i) Here, α + β = \(\frac { 1 }{ 4 }\) and α.β = – 1

Thus the polynomial formed

= x^{2} – (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 4x^{2} – 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

= x^{2} – (Sum of zeroes) x + Product of zeroes

= x^{2} – \(\sqrt { 2 }\) 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

3x^{2} – \(3\sqrt { 2 }x\) + 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

= x^{2} – (Sum of zeroes) x + Product of zeroes

= x^{2} – (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 ax^{3} + bx^{2} + cx + d

⇒ x^{3} + \(\frac { b }{ a }\)x^{2} + \(\frac { c }{ a }\)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

x^{3} + (–2) x^{2} + (–7)x + 14

⇒ x^{3} – 2x^{2} – 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 ax^{3} + bx^{2} + cx + d

⇒ x^{3} + \(\frac { b }{ a }\)x^{2} + \(\frac { c }{ a }\)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

x^{3} – (0) x^{2} + (–7)x + (–6)

⇒ x^{3} – 7x + 6

**Example 8: **If α and β are the zeroes of the polynomials ax^{2} + 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 ax^{2} + 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

x^{2} – (sum of zeroes) x + Product of zeroes

\(\Rightarrow {{\text{x}}^{2}}-\left( \frac{-b}{c} \right)\text{x}+\left( \frac{a}{c} \right)\)

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

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

⇒ cx^{2} + bx + a