**Introduction**

We use the standard form of a quadratic polynomial to find the Relationship between Zeroes and Coefficients of a Quadratic Polynomial. We know the standard form of quadratic polynomial is f(x) = ax^{2 }+ bx + c.

**Derivation of Formula**

Let α and β are the two zeroes of this polynomial. Then (x – α) and (x – β) will be factors of f(x).

So, for constant k we can write

f(x) = k(x – α) (x – β)

ax^{2 }+ bx + c = k{x^{2} – (α + β)x + αβ}

ax^{2 }+ bx + c = kx^{2} – k(α + β)x + kαβ

by comparing, a = k, b= – k(α + β), c = kαβ

∵ b= – k(α + β) and c = kαβ

α + β = b/-k and αβ = c/k

∵ a = k

α + β = b/-a or -b/a and αβ = c/a

So, for quadratic polynomial f(x) = ax^{2} + bx + c

**sum of zeroes (α + β) = -b/a = -(coefficient of x)/(coefficient of x ^{2})**

**product of zeroes (αβ) = c/a = (constant term)/(coefficient of x ^{2})**

**Some Examples**

**Example – 1)** find the zeroes of the quadratic polynomial 3x^{2} + 5x – 2 and verify the relationship between the zeroes and the coefficients.

**Solution – **let f(x) = 3x^{2} + 5x – 2

Now f(x) = 0

3x^{2} + 5x – 2 = 0

3x^{2} + 6x – x – 2 = 0

3x(x + 2) – 1(x + 2) = 0

(x + 2)(3x – 1) = 0

x + 2 = 0 and 3x – 1 = 0

**x = – 2** and **x = ⅓**

So, the zeroes of polynomial f(x) are x = – 2 and x = ⅓

Now sum of the zeroes = – 2 + ⅓ = (-6 + 1)/3 = **-5/3** = -(coefficient of x)/(coefficient of x^{2})

product of zeroes = – 2⨯⅓ = **-2/3** = (constant term)/(coefficient of x^{2})

So, the relationship between the zeroes and the coefficients of the quadratic polynomial is verified.

**Example – 2)** find a quadratic polynomial, whose sum and product of zeroes are -6 and 5 respectively.

**Solution –** let α and β are the zeroes of a quadratic polynomial.

For any constant k the quadratic polynomial will be

k{x^{2} – (α + β)x + αβ}

In question, sum of zeroes (α + β) = – 6

product of zeroes (αβ) = 5

putting the values,

k{x^{2} – (- 6)x + 5}

k{**x ^{2} + 6x + 5**} [Where k = Constant]

So, the required quadratic polynomial is x^{2} + 6x + 5. **Ans.**

**Example – 3)** find all the zeroes of polynomial f(x) = x^{3} + 13x^{2} + 32x + 20, if it’s one zero is – 2.

**Solution – **Here, – 2 is zero so the factor will be (x + 2). It is one factor of f(x).

To find other zeroes we will divide polynomial f(x) by factor (x + 2).

From division algorithm, quotient x^{2 }+ 11x + 10 will be its factor because remainder is 0.

Now Dividend = Divisor ⨯ Quotient + Remainder

x^{3} + 13x^{2} + 32x + 20 = (x + 2)⨯(x^{2 }+ 11x + 10) + 0

x^{3} + 13x^{2} + 32x + 20 = (x + 2)⨯{ x^{2} + 10x + x + 10}

x^{3} + 13x^{2} + 32x + 20 = (x + 2)⨯{ x(x + 10) + 1(x + 10)}

x^{3} + 13x^{2} + 32x + 20 = (x + 2)⨯(x + 10)(x + 1)

To find zeroes, f(x) = 0

(x + 2)(x + 10)(x + 1) = 0

**x = -2**, **x = -10** and **x = -1**

So, all the zeroes of polynomial f(x) are -2, -10, and -1. **Ans.**

**Relationship Between Zeroes and Coefficients of a Quadratic Polynomial Class 10 ^{th} in Hindi**

**More About Relationship Between Zeroes and Coefficients of a Quadratic Polynomial**

Deepika maliVery nice explained. written in detail about this topic..good job.

mitacademyssirohi2021great compliment for us. tq