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Polynomials

    Relationship between Zeroes and Coefficients of a Polynomial

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10th class > Polynomials > Relationship between Zeroes and Coefficients of a Polynomial

Relationship between Zeroes and Coefficients of a Polynomial

You have already seen that zero of a linear polynomial ax + b is −ba . We will now try to answer the question raised in the previous section regarding the relationship between zeroes and coefficients of a quadratic polynomial. For this, let us take a quadratic polynomial, say p(x) = 2x2-8x + 6. Previously, you have learnt how to factorise quadratic polynomials by splitting the middle term. So, here we need to split the middle term ‘– 8x’ as a sum of two terms, whose product is 6 × 2x2 = x2

So, we write 2x2 -8x + 6 = 2x2 -6x -2x + 6 = 2x(x – 3) – 2(x – 3)= (2x – 2)(x – 3) = 2(x – 1)(x – 3)

So, the value of p(x) = 2x2 -8x + 6 = is zero when x – 1 = or x – 3 =

i.e., when x = 1 or x = 3. So, the zeroes of 2x2 -8x + 6 are 1 and 3. Observe that :

Sum of its zeroes = 1 + 3 = = -(-8)2 = −Coefficient of xCoefficient ofx2

Product of its zeroes = 1 × 3 = = 62 = Constant termCoefficient ofx2

In general, α and β are the zeroes of the quadratic polynomial p(x) = ax2 + bx +c a ≠ 0, then you know that x – α and x – β are the factors of p(x). Therefore,

ax2 + b x + c = k(x – α ) (x – β), where k is a constant

= k[x2– ( α+ β)x + αβ]

= kx2 - k( α+ β) x + kαβ

Comparing the coefficients of x2, x and constant terms on both the sides, we get

a = k, b = – k(α + β ) and c = kαβ

This gives α + β = (since a=k)

αβ = (since a=k)

Let us take one more quadratic polynomial, say, p(x) = 3x2+5x−2.

By the method of splitting the middle term,

Since p(x) can have atmost three zeroes, these are the zeores of 2x3−5x2−14x+8.

4,-2 and 12 are the zeroes of that equation

Sum of zeros = 4 + (-2) + 12 = 52 = = .

Product of zeros = 4×−2×12= -4 = = .

For cubic polynomials we can also get one more identity. Mix the zeros two at a time and we get the following:

4×−2+4×12+−2×12= -8 + 2 - 1 = -7 = = .

In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d, then

α+β+γ=−ba

αβ+βγ+γα=ca

αβγ=−da

2. Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.

We have x2 + 7x + 10 = (x + )(x + )
So, the value of x2 + 7x + 10 = is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = or x = . Therefore, the zeroes of x2+ 7x + 10 are and .
Now, sum of zeroes = -2 + (-5) =
We get, sum of zeroes = -2 + (-5) = -7 = −71 = −Coefficient of xCoefficient ofx2
And the product of zeroes = (-2) × (-5) =
We get product of zeroes = (-2) × (-5) = 10 = 101 = Constant termCoefficient ofx2

3. Find the zeroes of the polynomial x2 -3 and verify the relationship between the zeroes and the coefficients.

Recall the identity a2 - b2 = (a – b)(a + b). Using it, we can write:
x2 -3 = (x - √3) (x + √3)
So, the value of x2 -3 is zero when x = or x =
Therefore, the zeroes of x2-3 are √3 and -√3
Now, the sum of zeroes = √3 - √3 = = −CoefficientofxCoefficientofx2
And the product of zeroes = (√3)(-√3) = = ConstanttermCoefficientofx2

4. Verify that 3, –1, −13 are the zeroes of the cubic polynomial p(x) = 3x3-5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients.

Verify the relation and the zeroes

  • Comparing the given polynomial with ax3 + bx2 + cx + d. we get a = , b = , c = , d =
  • Further: p(3) = 3 × 33 - (5 × 32) - (11 × 3) -3 = – – – (substitute the value of '3' in given eq)
  • Hence we get the answer is
  • p(-1)= 3 × −13 – 5 × −12– 11 × (–1) – 3 = – – + – (substitute the value of '-1' in given eq)
  • Hence we get the answer is
  • p(−13) = 3 × −133 - 5 × −132 - 11 × −13 - 3 = + + + (substitute the value of −13 in given eq)
  • Hence we get the answer is −23 + 23 =
  • Therefore, 3, –1 and −13 are the zeroes of 3x3-5x2 – 11x – 3
  • So, we take α = 3, β= -1, γ = −13
  • Now, α + β + γ = 3 + (-1) + (−13) = 2 - 13 =
  • αβ + βγ + γα = 3 × (-1) + (-1) × (−13) + (−13) × 3 = 3 + 13 - 1 =
  • αβγ = 3 × (-1) × (−13) =
  • We found the required answers.