Class 10 Polynomial
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Class 10 Chapter 2 - POLYNOMIAL

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` .
 
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