Polynomials

 

Chapter 2

Polynomial 

Polynomial

An algebraic expression can have exponents that are rational numbers. However, a polynomial is an algebraic expression in which the exponent on any variable is a whole number.

5x3+3x+1 is an example of a polynomial. It is an algebraic expression as well.

2x+3√x is an algebraic expression, but not a polynomial. – since the exponent on x is 1/2 which is not a whole number.

Degree of a Polynomial

For a polynomial in one variable – the highest exponent on the variable in a polynomial is the degree of the polynomial.

Example: The degree of the polynomial x2+2x+3 is 2, as the highest power of x in the given expression is x2.

Types Of Polynomials

Polynomials can be classified based on:
a) Number of terms
b) Degree of the polynomial.

Types of polynomials based on the number of terms

a) Monomial – A polynomial with just one term. Example: 2x, 6x2, 9xy

b) Binomial – A polynomial with two terms. Example: 4x2+x, 5x+4

a) Trinomial – A polynomial with three terms. Example: x2+3x+4

Types of Polynomials based on Degree

Linear Polynomial

A polynomial whose degree is one is called a linear polynomial.
For example, 2x+1 is a linear polynomial.

Quadratic Polynomial

A polynomial of degree two is called a quadratic polynomial.
For example, 3x2+8x+5 is a quadratic polynomial.

Cubic Polynomial

A polynomial of degree three is called a cubic polynomial.
For example, 2x3+5x2+9x+15 is a cubic polynomial.

Zeroes of a Polynomial

A zero of a polynomial p(x) is the value of x for which the value of p(x) is 0. If k is a zero of p(x), then p(k)=0.
For example, consider a polynomial p(x)=x2−3x+2.
When x=1, the value of p(x) will be equal to
p(1)=12−3×1+2
=1−3+2
=0

Since p(x)=0 at x=1, we say that 1 is a zero of the polynomial x2−3x+2

Factorisation of Polynomials

Quadratic polynomials can be factorized by splitting the middle term.

For example, consider the polynomial 2x2−5x+3

Splitting the middle term:

The middle term in the polynomial 2x2−5x+3 is -5x. This must be expressed as a sum of two terms such that the product of their coefficients is equal to the product of 2 and 3 (coefficient of x2 and the constant term)

−5 can be expressed as (−2)+(−3), as −2×−3=6=2×3

Thus, 2x2−5x+3=2x2−2x−3x+3

Now, identify the common factors in individual groups

2x2−2x−3x+3=2x(x−1)−3(x−1)

Taking (x−1) as the common factor, this can be expressed as:

2x(x−1)−3(x−1)=(x−1)(2x−3)

Relationship between Zeroes and Coefficients of a Polynomial

For Quadratic Polynomial:
If α and β are the roots of a quadratic polynomial ax2+bx+c, then,

α +  β = -b/a

Sum of zeroes = -coefficient of x /coefficient of x2

αβ = c/a

Product of zeroes = constant term / coefficient of x2

For Cubic Polynomial

If α,β and γ are the roots of a cubic polynomial ax3+bx2+cx+d, then

α+β+γ = -b/a

αβ +βγ +γα = c/a

αβγ = -d/a

Division Algorithm

To divide one polynomial by another, follow the steps given below.

Step 1: arrange the terms of the dividend and the divisor in the decreasing order of their degrees.

Step 2: To obtain the first term of the quotient, divide the highest degree term of the dividend by the highest degree term of the divisor  Then carry out the division process.

Step 3: The remainder from the previous division becomes the dividend for the next step. Repeat this process until the degree of the remainder is less than the degree of the divisor.

Algebraic Identities

1. (a+b)2=a2+2ab+b2
2. (a−b)2=a2−2ab+b2
3. (x+a)(x+b)=x2+(a+b)x+ab
4. a2−b2=(a+b)(a−b)
5. a3−b3=(a−b)(a2+ab+b2)
6. a3+b3=(a+b)(a2−ab+b2)
7. (a+b)3=a3+3a2b+3ab2+b3
8. (a−b)3=a3−3a2b+3ab2−b3

Exercises 2.1

1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Answer

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

Exercise 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u

(v) t² – 15

(vi) 3x² – x – 4

Answer

(i) x² – 2x – 8
= (x - 4) (x + 2)
The value of x² – 2x – 8 is zero when x - 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x² – 2x – 8 are 4 and -2.

Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x²

Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of x²

(ii) 4s² – 4s + 1
= (2s-1)²
The value of 4s² - 4s + 1 is zero when 2s - 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s² - 4s + 1 are 1/2 and 1/2.

Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s²
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s².

(iii) 6x² – 3 – 7x
= 6x² – 7x – 3
= (3x + 1) (2x - 3)
The value of 6x² - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2

Therefore, the zeroes of 6x² - 3 - 7x are -1/3 and 3/2.

Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x²
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x².

(iv) 4u² + 8u
= 4u² + 8u + 0
= 4u(u + 2)
The value of 4u² + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2
Therefore, the zeroes of 4u² + 8u are 0 and - 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u²
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u².

(v) t² – 15
= t² - 0.t - 15
= (t - √15) (t + √15)
The value of t² - 15 is zero when t - √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15
Therefore, the zeroes of t² - 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t²
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u².

(vi) 3x² – x – 4
= (3x - 4) (x + 1)
The value of 3x² – x – 4 is zero when 3x - 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3x² – x – 4 are 4/3 and -1.

Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x²

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x².

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4 , -1

(ii) √2 , 1/3 

(iii) 0, √5

(iv) 1,1 

(v) -1/4 ,1/4 

(vi) 4,1

Answer

(i) 1/4 , -1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 1/4 = -b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x² - x -4.

(ii) √2 , 1/3
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = √2 = 3√2/3 = -b/a
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3x² -3√2x +1.

(iii) 0, √5
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = -b/a
αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x² + √5.

(iv) 1, 1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 1 = 1/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x² - x +1.

(v) -1/4 ,1/4
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = -1/4 = -b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x² + x +1.

(vi) 4,1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 4 = 4/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x² - 4x +1.

Exercise 2.3

1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

Answer

(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2

Quotient = x-3 and remainder 7x - 9

(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

Quotient = x² + x - 3 and remainder 8

(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²

Quotient = -x² -2 and remainder -5x +10

2. Check whether the first polynomial is a factor of the second polynomial by dividing the
second polynomial by the first polynomial:

Answer

(i) t² – 3,  2t⁴ + 3t³ – 2t² – 9t – 12

t² – 3 exactly divides  2t⁴ + 3t³ – 2t² – 9t – 12 leaving no remainder. Hence, it is a factor of  2t⁴ + 3t³ – 2t² – 9t – 12.

(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

x² + 3x + 1 exactly divides 3x⁴ + 5x³ – 7x² + 2x + 2 leaving no remainder. Hence, it is factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

x³ – 3x + 1 didn't divides exactly x⁵ – 4x³ + x² + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of x⁵ – 4x³ + x² + 3x + 1.

3. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are √(5/3)

and - √(5/3).

Answer

p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5
Since the two zeroes are √(5/3) and - √(5/3).

We factorize x² + 2x + 1

= (x + 1)²

Therefore, its zero is given by x + 1 = 0

x = -1

As it has the term (x + 1)² , therefore, there will be 2 zeroes at x = - 1.
Hence, the zeroes of the given polynomial are √(5/3) and - √(5/3), - 1 and - 1.

4.  On dividing x³ - 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and 
-2x + 4, respectively. Find g(x).

Answer

Here in the given question,

Dividend = x³ - 3x² + x + 2

Quotient = x - 2

Remainder = -2x + 4

Divisor = g(x)

We know that,

Dividend = Quotient × Divisor + Remainder

⇒ x³ - 3x² + x + 2 = (x - 2) × g(x) + (-2x + 4)⇒ x³ - 3x² + x + 2 - (-2x + 4) = (x - 2) × g(x)
⇒ x³ - 3x² + 3x - 2 = (x - 2) × g(x)
⇒ g(x) =  (x³ - 3x² + 3x - 2)/(x - 2)

∴ g(x) = (x² - x + 1)

5.Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0

Answer

(i) Let us assume the division of 6x² + 2x + 2 by 2
Here, p(x) = 6x² + 2x + 2
g(x) = 2
q(x) = 3x² + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x² + 2x + 2 = 2x (3x² + x + 1)
Hence, division algorithm is satisfied.

(ii) Let us assume the division of x³+ x by x²,
Here, p(x) = x³ + x
g(x) = x²
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + x = (x² ) × x + x
x³ + x = x³ + x
Thus, the division algorithm is satisfied.

(iii) Let us assume the division of x³+ 1 by x².
Here, p(x) = x³ + 1
g(x) = x²
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + 1 = (x² ) × x + 1
x³ + 1 = x³ + 1
Thus, the division algorithm is satisfied.

Exercise 2.4 (Optional)

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x³ + x² - 5x + 2; 1/2, 1, -2
(ii) x³ - 4x² + 5x - 2; 2, 1, 1

Answer

(i) p(x) = 2x³ + x² - 5x + 2
Now for zeroes, putting the given value in x.

p(1/2) = 2(1/2)³ + (1/2)² - 5(1/2) + 2
= (2×1/8) + 1/4 - 5/2 + 2
= 1/4 + 1/4 - 5/2 + 2
= 1/2 - 5/2 + 2 = 0

p(1) = 2(1)³ + (1)² - 5(1) + 2
= (2×1) + 1 - 5 + 2
= 2 + 1 - 5 + 2 = 0

p(-2) = 2(-2)³ + (-2)² - 5(-2) + 2
= (2 × -8) + 4 + 10 + 2
= -16 + 16 = 0

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with ax³ + bx² + cx + d, we get a=2, b=1, c=-5, d=2
Also, α=1/2, β=1 and γ=-2
Now,
-b/a = α+β+γ
⇒ 1/2 = 1/2 + 1 - 2
⇒ 1/2 = 1/2

c/a = αβ+βγ+γα
⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)
⇒ -5/2 = 1/2 - 2 - 1
⇒ -5/2 = -5/2

-d/a = αβγ
⇒ -2/2 = (1/2 × 1 × -2)
⇒ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii)  p(x) = x³ - 4x² + 5x - 2
Now for zeroes, putting the given value in x.

p(2) = 2³ - 4(2)² + 5(2) - 2
= 8 - 16 + 10 - 2
= 0

p(1) = 1³ - 4(1)² + 5(1) - 2
= 1 - 4 + 5 - 2
= 0

p(1) = 1³ - 4(1)² + 5(1) - 2
= 1 - 4 + 5 - 2
= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with ax³ + bx² + cx + d, we get a=1, b=-4, c=5, d=-2
Also, α=2, β=1 and γ=1
Now,
-b/a = α+β+γ
⇒ 4/1 = 2 + 1 + 1
⇒ 4 = 4

c/a = αβ+βγ+γα
⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)
⇒ 5 = 2 + 1 + 2
⇒ 5 = 5

-d/a = αβγ
⇒ 2/1 = (2 × 1 × 1)
⇒ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.

Answer

Let the polynomial be ax³ + bx² + cx + d and the zeroes be α, β and γ
Then, α + β + γ = -(-2)/1 = 2 = -b/a
αβ + βγ + γα = -7 = -7/1 = c/a
αβγ = -14 = -14/1 = -d/a

∴ a = 1, b = -2, c = -7 and d = 14
So, one cubic polynomial which satisfy the given conditions will be x³ - 2x²  - 7x + 14

3. If the zeroes of the polynomial x³ – 3x² + x + 1 are a–b, a, a+b, find a and b.

Answer

Since, (a - b), a, (a + b) are the zeroes of the polynomial x³ – 3x² + x + 1.
Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

⇒ 3a = 3 ⇒ a =1

∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1
a² - ab + a² + ab + a² - b² = 1
⇒ 3a² - b² =1

Putting the value of a,

⇒ 3(1)² - b² = 1
⇒ 3 - b² = 1
⇒ b² = 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2

4. If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are  2±√3,  find other zeroes.

Answer

2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x⁴ – 6x³ – 26x² + 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x² - 4x + 4 = 3,
⇒ x² - 4x + 1= 0

Now, dividing p(x) by x² - 4x + 1

∴ p(x) = x⁴ - 6x³ - 26x² + 138x - 35
= (x² - 4x + 1) (x² - 2x - 35)
= (x² - 4x + 1) (x² - 7x + 5x - 35)
= (x² - 4x + 1) [x(x - 7) + 5 (x - 7)]
= (x² - 4x + 1) (x + 5) (x - 7)

∴ (x + 5) and (x - 7) are other factors of p(x).
∴ - 5 and 7 are other zeroes of the given polynomial.

5. If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.

Answer

On dividing x⁴ – 6x³ + 16x² – 25x + 10 by x² – 2x + k

∴ Remainder = (2k - 9)x - (8 - k)k + 10 

But the remainder is given as x+ a. 

On comparing their coefficients,

2k - 9 = 1

⇒ k = 10 

⇒ k = 5 and,

-(8-k)k + 10 = a

⇒ a = -(8 - 5)5 + 10 =- 15 + 10 = -5 

Hence, k = 5 and a = -5 

Find the zeroes of 2x² - 8x + 6.

We have:
2x² - 8x + 6 = 2x² - 6x - 2x + 6
= 2x (x - 3) - 2 (x - 3)
= (2x - 2) (x - 3)
= 2 (x - 1) (x - 3)
For 2x² - 8x + 6 to be zero,
Either, x - 1 = 0 ⇒ x = 1
or x - 3 = 0 ⇒ x = 3
∴ The zeroes of 2x² - 8x + 6 are 1 and 3.