**Relationship Between Zeros And Coefficients Of A Polynomial**

Consider quadratic polynomial

P(x) = 2x^{2} – 16x + 30.

Now, 2x^{2} – 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 = =

Product of the zeros = 3 × 5 = 15 = =

So if ax^{2} + bx + c, a ≠ 0 is a quadratic polynomial and α, β are two zeros of polynomial then

In general, it can be proved that if α, β, γ are the zeros of a cubic polynomial ax^{3} + bx^{2} + cx + d, then

Note: , and 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 6x^{2} – 13x + 6 and verify the relation between the zeros and its coefficients.

**Sol.** We have, 6x^{2} – 13x + 6 = 6×2 – 4x – 9x + 6

= 2x (3x – 2) –3 (3x – 2)

= (3x – 2) (2x – 3)

So, the value of 6x^{2} – 13x + 6 is 0, when

(3x – 2) = 0 or (2x – 3) = 0 i.e.,

When x = or

Therefore, the zeros of 6x^{2} – 13x + 6 are

and

Sum of the zeros

= + = = =

Product of the zeros

= × = =

**Example 2: **Find the zeros of the quadratic polynomial 4×2 – 9 and verify the relation between the zeros and its coefficients.

**Sol.** We have,

4x^{2} – 9 = (2x)^{2} – 3^{2} = (2x – 3) (2x + 3)

So, the value of 4x^{2} – 9 is 0, when

2x – 3 = 0 or 2x + 3 = 0

i.e., when x = or x = .

Therefore, the zeros of 4x^{2} – 9 are & .

Sum of the zeros

= = 0 = =

Product of the zeros

= × = =

**Example 3: **Find the zeros of the quadratic polynomial 9x^{2} – 5 and verify the relation between the zeros and its coefficients.

**Sol.** We have,

9x^{2} – 5 = (3x)^{2} – (√5)^{2} = (3x – √5) (3x + √5)

So, the value of 9x^{2} – 5 is 0,

when 3x – √5 = 0 or 3x + √5 = 0

i.e., when x = or x = .

Sum of the zeros

= = 0 = =

Product of the zeros

= × = =

**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 ax^{2} + bx + c.

Therefore, (x – α), (x – β) are the factors of the polynomial ax^{2} + bx + c.

⇒ ax^{2} + bx + c = 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 …(1)

Comparing the coefficients of x^{2}, x and constant terms of (1) on both sides, we get

a = k, b = – k (α + β) and c = kαβ

⇒ α + β = and αβ =

α + β = and αβ = [∵ k = a]

Sum of the zeros = =

Product of the zeros = =

**Example 5: **Prove relation between the zeros and the coefficient of the quadratic polynomial

ax^{2} + bx + c.

**Sol.** Let a and b be the zeros of the polynomial ax^{2} + bx + c

α = ….(1)

β = ….(2)

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

α + β = +

= = =

Hence, sum of the zeros of the polynomial

ax^{2} + bx + c is

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

αβ = ×

=

= =

=

Hence, product of zeros =

**Example 6: **find the zeroes of the quadratic polynomial x^{2} – 2x – 8 and verify a relationship between zeroes and its coefficients.

**Sol.** x^{2} – 2x – 8 = x^{2} – 4x + 2x – 8

= x (x – 4) + 2 (x – 4) = (x – 4) (x + 2)

So, the value of x^{2} – 2x – 8 is zero when

x – 4 = 0 or x + 2 = 0 i.e., when x = 4 or x = – 2.

So, the zeroes of x^{2} – 2x – 8 are 4, – 2.

Sum of the zeroes

= 4 – 2 = 2 = =

Product of the zeroes

= 4 (–2) = –8 = =

**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. 2x^{3} + x^{2} – 5x + 2 ; , 1, – 2

**Sol. ** Here, the polynomial p(x) is 2x^{3} + x^{2} – 5x + 2

Value of the polynomial 2x^{3} + x^{2} – 5x + 2

when x = 1/2

= = = 0

So, 1/2 is a zero of p(x).

On putting x = 1 in the cubic polynomial

2x^{3} + x^{2} – 5x + 2

= 2(1)3 + (1)2 –¬ 5(1) + 2 = 2 + 1 – 5 + 2 = 0

On putting x = – 2 in the cubic polynomial

2x^{3} + x^{2} – 5x + 2

= 2(–2)3 + (–2)2 – 5 (–2) + 2

= – 16 + 4 + 10 + 2 = 0

Hence, , 1, – 2 are the zeroes of the given polynomial.

Sum of the zeroes of p(x)

= + 1 – 2 = =

Sum of the products of two zeroes taken at a time

= × 1 + × (–2) + 1 × (–2)

= – 1 – 2 = =

Product of all the three zeroes

= × (1) × (–2) = –1

= =