Class 10 Polynomial

### Topic covered

♦ Relationship between Zeroes and Coefficients of a Polynomial

### Relationship between Zeroes and Coefficients of a Polynomial

We will now try to answer the question raised regarding the relationship between zeroes
and coefficients of a quadratic polynomial.

color(red)(=>"For this, let us take a quadratic polynomial,") say color(blue)(p(x) = 2x^2 – 8x + 6).

In Class IX, 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 × 2x^2 = 12x^2.

So, we write

color(blue)(2x^2 – 8x + 6) = 2x^2 – 6x – 2x + 6 = 2x(x – 3) – 2(x – 3)

" " = (2x – 2)(x – 3) = 2(x – 1)(x – 3)

So, the value of p(x) = 2x^2 – 8x + 6 is zero when x – 1 = 0 or x – 3 = 0,

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

color(blue)("Sum of its zeroes" = 1+3 =4 = (-8)/2 = ( - ( text (Coefficient of x ) ) )/( text (coefficient of) x^2 ) )

color(blue)("Product of its zeroes" = 1 xx 3 = 3 = 6/2 = ( text (Constant term ) )/( text (Coefficient of) x^2 ) )

color(red)(=>"Let us take one more quadratic polynomial"), say, color(blue)(p(x) = 3x^2 + 5x – 2)

By the method of splitting the middle term,

color(blue)(3x^2 + 5x – 2 = 3x^2 + 6x – x – 2 = 3x(x + 2) –1(x + 2))

" " = (3x – 1)(x + 2)

Hence, the value of 3x^2 + 5x – 2 is zero when either 3x – 1 = 0 or x + 2 = 0, i.e.,

when x = 1/3 or x = -2 So , the zeroes of 3x^2 + 5x – 2 are 1/3 and -2 . Observe that :

color(blue)("Sum of its zeroes" = 1/3 + (-2) = (-5)/3 = ( ( text (Coefficient of x ) ) )/( text (Coefficient of )x^2) )

color(blue)("Product of its zeroes" = 1/3 xx (-2) = (-2)/3 = ( text ( Constant term ) )/( text (Coefficient of )x^2) )

In general, if α and β are the zeroes of the quadratic polynomial color(red)(p(x) = ax^2 + bx + c),

a ≠ 0, then you know that x – α and x – β are the factors of p(x). Therefore,

color(red)(ax^2 + bx + c = k(x – α) (x – β)), where k is a constant

" " = k [ x^2 – (α + β)x + α β]

" " = kx^2 – k(α + β)x + k α β

Comparing the coefficients of x^2, x and constant terms on both the sides, we get

a = k," " b = – k(α + β) and  " " c = kαβ.

This gives  color(blue)(α + β = (-b)/a)

color(blue)( α β = c/a)

α,\ \ β are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively. We will use later one more letter ‘γ’ pronounced as ‘gamma’.

i.e., color(red)("sum of zeroes "= α + β = - b/a = ( - text ( Coefficient of x ) )/( text (Coefficient of) x^2) ),

color(red)("product of zeroes "= αβ = c/a = ( text (Constant term ) )/( text (Coefficient of )x^2 ) ) .

Q 3139056812

Find the zeroes of the quadratic polynomial x^2 + 7x + 10, and verify the
relationship between the zeroes and the coefficients.
Class 12 Chapter 2 Example 2
Solution:

We have

x^2 + 7x + 10 = (x + 2)(x + 5)

So, the value of x^2 + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = – 2 or
x = –5. Therefore, the zeroes of x^2 + 7x + 10 are – 2 and – 5. Now,

sum of zeroes  = -2 + (-5) = - (7 ) = ( - (7 ) )/1 = ( - ( text (Coefficient of x) ) )/( text (Coefficient of x^2) ) ,

product of zeroes = (-2) xx (-5) = 10 = 10/1 = ( text (Constant term ) )/( text (Coefficient of x^2 ) ) .
Q 3159056814

Find the zeroes of the polynomial x^2 – 3 and verify the relationship
between the zeroes and the coefficients.
Class 10 Chapter 2 Example 3
Solution:

Recall the identity a^2 – b^2 = (a – b)(a + b). Using it, we can write:

 x^2 -3 = (x- sqrt 3 ) ( x+ sqrt 3 )

So, the value of x^2 – 3 is zero when x = sqrt 3 or x = - sqrt 3 .

Therefore, the zeroes of x^2 – 3 are sqrt 3 and  - sqrt 3 .

Now,

sum of zeroes = sqrt 3 - sqrt 3 = 0 = ( - (text (Coefficient of x) ) )/( text (Coefficient of x^2) ) ,

product of zeroes = ( sqrt 3 ) ( sqrt 3 ) = = - 3 = (-3)/1 = ( text (Constant term) )/( text ( Coefficient of) x^2 )
Q 3119156910

Find a quadratic polynomial, the sum and product of whose zeroes are
–3 and 2, respectively.
Class 10 Chapter 2 Example 4
Solution:

Let the quadratic polynomial be ax^2 + bx + c , and its zeroes be α and β.
We have

 α + β = -3 = (-b)/a ,

and  αβ =2 = c/a

If a = 1, then b = 3 and c = 2.

So, one quadratic polynomial which fits the given conditions is x^2 + 3x + 2.

You can check that any other quadratic polynomial that fits these conditions will
be of the form k(x^2 + 3x + 2), where k is real.

Let us now look at cubic polynomials. Do you think a similar relation holds
between the zeroes of a cubic polynomial and its coefficients?
Let us consider p(x) = 2x^3 – 5x^2 – 14x + 8.

You can check that p(x) = 0 for x = 4, – 2, 1/2 . Since p(x) can have atmost three
zeroes, these are the zeores of 2x^3 – 5x^2 – 14x + 8. Now,

sum of the zeroes = 4+ (-2 ) + 1/2 = 5/2 = (- (-5) )/2 = ( - text (Coefficient of )x^2 )/( text (Coefficient of ) x^3 ),

product of the zeroes = 4 xx (-2) xx 1/2 = -4 = (-8)/2 = ( - text (Constant term ) )/( text ( Coefficient of) x^3 )

However, there is one more relationship here. Consider the sum of the products
of the zeroes taken two at a time. We have

 { 4 xx (-2) } + { (-2) xx 1/2 } + { 1/2 xx 4 }

= - 8 - 1 +2 = -7 = (-14)/2 = ( text (Coefficient of x) )/( text (Coefficient of ) x^3 ).

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

α + β + γ = (-b)/a ,

 αβ + βγ + γα = c/a ,

 α β γ = (-d)/a .
Let us consider an example.
Q 3119167010

Verify that 3, –1, - 1/3 are the zeroes of the cubic polynomial
p(x) = 3x^3 – 5x^2 – 11x – 3, and then verify the relationship between the zeroes and the
coefficients.
Class 10 Chapter 2 Example 5
Solution:

Comparing the given polynomial with ax^3 + bx^2 + cx + d, we get

a = 3, b = – 5, c = –11, d = – 3. Further

p(3) = 3 × 3^3 – (5 × 3^2) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,

p(–1) = 3 × (–1)^3 – 5 × (–1)^2 – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,

p ( -1/3) = 3 xx (-1/3)^3 -5 xx ( -1/3)^2 -11 xx (-1/3 ) -3 ,

 = -1/9 -5/9 + 11/3 -3 = -2/3 + 2/3 = 0

Therefore, 3, –1 and  - 1/3 are the zeroes of 3x^3 – 5x^2 – 11x – 3.

So, we take α = 3, β = –1 and γ = -1/3 .

Now,

 α + β + γ = 3 + (-1) + (-1/3) = 2 -1/3 = 5/3 = ( - (-5) )/3 = (-b)/a ,

αβ+ βγ + γα = 3 × (-1) + (-1) xx (-1/3) + (-1/3 ) xx 3 = -3 +1/3 -1 = (-11)/3 = c/a ,

 αβγ = 3 × (-1) xx (-1/3) =1 = (- (-3) )/3 = (-d)/a .