Polynomial Expression in Nature and Technology (Class- 9, Experience- 4)

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• MCQ Test
• Exercise Solution

[Exercise- page 112]

E.4.1 1. Form polynomial expressions from three real life examples.

Solution:

Form polynomial expressions from three real life examples.

1. Cost of a meal: Consider going to a restaurant where the cost of a meal includes a fixed base price (for example, the cost of a meal without any additional items) and additional charges for each item added (like appetizers, drinks, desserts, etc.). Let's denote the base price by b and the additional charge for each item by c. If you order n additional items, the total cost of your meal can be represented as: 

Polynomial expression, Total cost, C= b + cn, where C is the total cost of the meal and n is the number of additional items ordered.

2. Area of a Rectangle: Let's say we have a rectangle with length (L) and width (W). The area (A) of the rectangle can be expressed as the product of its length and width. 

Polynomial expression, Area, A=LW

3. Distance Traveled by a Moving Object: Imagine a car traveling at a constant speed of v meters per second. The distance traveled by the car over time t seconds can be expressed as the product of its speed (v) and time (t). Polynomial expression, distance, D = vt 

In this example, the polynomial expression represents the relationship between the distance traveled by the car and the time it has been traveling, assuming a constant speed.

E.4.2 Give examples of polynomial expressions according to the following instructions. 

i) one variable, three dimensional, binomial 

ii) one variable, three dimensional, quadrimonial 

iii) two variables, three dimensional, binomial 

iv) two variables, three dimensional, trinomial 

v) four variables, cyclic, four dimensional

Solution:

i) One variable, three dimensional, binomial: 

    2x³−5x

ii) One variable, three dimensional, quadrinomial: 

     3x³+2x²−5x+7

iii) Two variables, three dimensional, binomial: 

      2xy³−5y

iv) Two variables, three dimensional, trinomial: 

      3x³y−2xy²+7xy

v) Four variables, cyclic, four dimensional:

    2wxyz+3wx³z²−5xy²z+7w²y²

E.4.3 Give example: 

i) homogeneous, symmetric, cyclic polynomial 

ii) homogeneous, symmetric, polynomial; not cyclic 

iii) homogeneous, cyclic polynomial; not symmetric 

iv) symmetric, cyclic polynomial; not homogeneous

Solution:

i) homogeneous, symmetric, cyclic polynomial

Here's an example that satisfies all three conditions:

f(x,y,z) = x³ + y³ + z³ − 3xyz

Explanation:

Homogeneous: Each term has a degree of 3.

Symmetric: The terms are symmetric in x, y, and z, meaning if you swap the variables, the expression remains the same.

Cyclic: The expression remains unchanged under cyclic permutations of x, y, and z. For example, if we cyclically permute x, y, and z as y→z→x, the expression remains the same.

Let's break down each of the terms:

Homogeneous Polynomial: A polynomial is homogeneous if all its terms have the same total degree.

Symmetric Polynomial: A polynomial is symmetric if it remains unchanged when the variables are permuted.

Cyclic Polynomial: A polynomial is cyclic if it remains unchanged under cyclic permutations of its variables.

ii) homogeneous, symmetric, polynomial; not cyclic

Here's an example of a polynomial that is homogeneous and 

symmetric but not cyclic:

f(x,y,z) = x² + y² + z² − xy − yz − xz

Explanation:

Homogeneous: Each term has a degree of 2, making the polynomial homogeneous.

Symmetric: The terms are symmetric in x, y, and z, meaning if you swap the variables, the expression remains the same.

Not Cyclic: Although the polynomial is symmetric, it is not cyclic. Cyclic polynomials remain unchanged under cyclic permutations of the variables, but in this case, permuting the variables does not preserve the expression. For example, if we cyclically permute x, y, and z as y→z→x, the expression changes.

Therefore, this polynomial is homogeneous and symmetric but not cyclic.

iii) homogeneous, cyclic polynomial; not symmetric 

Here's an example of a polynomial that is homogeneous, cyclic, but not symmetric:

f(x,y,z) = x³ + y³ + z³ − 3xyz

Explanation:

Homogeneous: Each term has a degree of 3, making the polynomial homogeneous.

Cyclic: This polynomial is cyclic because it remains unchanged under cyclic permutations of the variables. For example, if we cyclically permute x, y, and z as y→z→x, the expression remains the same.

Not Symmetric: Although the polynomial is homogeneous and cyclic, it is not symmetric. A symmetric polynomial remains unchanged when variables are permuted, but in this case, permuting the variables does not preserve the expression. For example, swapping x and y as y→x changes the expression.

Therefore, this polynomial is homogeneous and cyclic but not symmetric.

iv) symmetric, cyclic polynomial; not homogeneous

Here's an example of a polynomial that is symmetric, cyclic, but not homogeneous:

f(x,y,z) = x² + y² + z² − xy − yz − zx

Explanation:

Symmetric: The terms are symmetric in x, y, and z, meaning if you swap the variables, the expression remains the same.

Cyclic: This polynomial is cyclic because it remains unchanged under cyclic permutations of the variables. For example, if we cyclically permute x, y, and z as y→z→x, the expression remains the same.

Not Homogeneous: The degrees of the terms in this polynomial are not all the same. While some terms like x², y², and z² have a degree of 2, the terms involving products of variables like xy, yz, and zx have a degree of 1.

Therefore, this polynomial is symmetric and cyclic but not homogeneous.

E.4.4 i) Divide x⁴ - 3x² + 1  by  2x² - 3;

ii) Divide 5x³ - 3x - 2  by  3x - 2 and find out the remainder. Prove the appropriateness of the remainder through the remainder theorem.

Solution:

i) Divide x⁴ - 3x² + 1 by 2x² - 3

2x²-3) x⁴ - 3x² + 1 ( 12x² – 34

           -(x⁴ – 32x²)

_____________________

                   -32x² + 1

                 -(-32x² + 94)

_____________________

                              -54

∵ Answer:   12x² – 34 –   5/42x²-3

ii) Divide 5x³ - 3x - 2 by 3x - 2 and find out the remainder. Prove the appropriateness of the remainder through the remainder theorem.

3x – 2  ) 5x³ – 3x – 2 (  53x² + 10 9x –  727

            – (5x³ – 10 3x²)

_____________________________

               10 3x² – 3x

           -( 10 3x² – 20 9x)

______________________________

                       -79x – 2

                    -(-79x + 1427)

________________________________

                               -68 27

∵ remainder   -68 27

through the remainder theorem, Proving the appropriateness of the remainder:

Here, P(x) = 5x³ – 3x – 2

and 3x – 2, P(x) is a factor.

then, x = 23 then, find the value of P(x).

P(23) = 5(23)3 – 3(23) – 2

        = 5.827 – 2 – 2

        = 4027 – 4

        =40 - 108    27

        = -68 27

Here, Remainders are same [proved]

E.4.5 Which of the following expressions are real prime expressions? Factor the expressions that are not real prime expressions. 

i)  x² - 5x -14 

ii)  x² - 5x + 2 

iii)  2x² + 3x + 1 

iv)  3x² + 4x - 1

Solution:

i) x² - 5x -14 

To determine whether x^2−5x−14 is a prime expression, we can check if it can be factored into linear factors (factors of the form (x−a)). If it can be factored, then it's not prime. If it cannot be factored, then it's prime.

Let's factor x² − 5x − 14:

We're looking for two numbers that multiply to give us −14 and add to give us −5. These numbers are −7 and 2.

So, we can rewrite x² − 5x − 14 as:

    x² − 7x + 2x − 14

= (x² − 7x) + (2x − 14)

= x(x − 7) + 2(x − 7)

= (x − 7)(x + 2)

Therefore, x² − 5x − 14 is not a prime expression, as it can be factored into (x − 7)(x + 2).

ii) x² - 5x + 2 

To determine if x² − 5x + 2 is a prime expression, we need to check if it can be factored into linear factors. If it can be factored, then it's not prime. If it cannot be factored, then it's prime.

Let's attempt to factor x² −5x + 2:

We're looking for two numbers that multiply to give us 2 and add to give us −5. These numbers are −1 and −2.

So, we can rewrite x² − 5x + 2 as:

    x² − 2x − 3x + 2

= (x² − 2x) + (−3x + 2)

= x(x − 2) −1(3x − 2)

= (x − 2)(x − 1)

Therefore, x² − 5x + 2 is not a prime expression, as it can be factored into (x − 2)(x − 1).

iii) 2x² + 3x + 1 

To determine if 2x² + 3x + 1 is a prime expression, we need to check if it can be factored into linear factors. If it can be factored, then it's not prime. If it cannot be factored, then it's prime.

Let's attempt to factor 2x² + 3x + 1:

This expression is a quadratic trinomial, and we can check if it can be factored using the quadratic formula or by trying to find two numbers that multiply to give 2 (the coefficient of x²) and add to give 3 (the coefficient of x).

Using the quadratic formula:

x= -b±√ b² - 4ac         2a

where a=2, b=3, and c=1, we get:

x= -3±√ 3² - 4(2)(1)          2(2)

x= -3±√ 9 - 8        4

x= -3±√ 1     4

x= -3 ± 1    4

So, the roots are:

x= -3 + 1    4 = -12

x= -3 - 1    4 = -1

Now, we can express 2x² + 3x + 1 as:

2x² + 3x + 1= 2(x + 21)(x + 1)

Therefore, 2x² + 3x + 1 is not a prime expression, as it can be factored into 2(x + 21)(x + 1).

iv) 3x² + 4x – 1

To determine if 3x² +4x − 1 is a prime expression, we need to check if it can be factored into linear factors. If it can be factored, then it's not prime. If it cannot be factored, then it's prime.

Let's attempt to factor 3x² + 4x − 1:

This expression is a quadratic trinomial, and we can try to factor it by finding two numbers that multiply to give 3 × (−1) = −3 (the product of the coefficients of x² and the constant term) and add to give 4 (the coefficient of x).

The two numbers that satisfy these conditions are 3 and −1.

So, we can rewrite 3x² + 4x − 1 as:

    3x² + 3x + x − 1

= 3x(x + 1) + 1(x + 1)

= (3x + 1)(x + 1)

Therefore, 3x² + 4x − 1 is not a prime expression, as it can be factored into (3x + 1)(x + 1).

E.4.6 Factor the following expressions. 

i)  x³ - 5x + 4 

ii)  x³ - 3x² + 3x - 2 

iii)  x⁵ - 16xy⁴

Solution:

i) x³ - 5x + 4

Lets say, P(x) = x³ - 5x + 4

Here, if x = 1,

P(1) = 1³ - 5.1 + 4 = 1 – 5 + 4 = 0

Therefore, (x-1) is a factor of x³ - 5x + 4

So,

    x³ - 5x + 4

= x²(x-1) + x(x-1) - 4(x-1)

= (x-1)(x² + x - 4) [Answer]

ii) x³ - 3x² + 3x - 2

Lets say, P(x) = x³ - 3x² + 3x - 2

Here, if x = 2,

P(2) = 2³ – 3.2² + 3.2 – 2 

        = 8 – 12 + 6 – 2 

        = 14 – 14 

        = 0

Therefore, (x-2) is a factor of x³ - 3x² + 3x - 2

So,
    x³ - 3x² + 3x - 2

= x²(x-2) - x(x-2) + 1(x-2)

= (x-2)(x² - x + 1) [Answer]

iii) x⁵ - 16xy⁴

    x⁵ - 16xy⁴

= x(x⁴ - 16y⁴)

= x{x⁴ - (2y)⁴}

= x[{(x²)² - {(2y)²}²]

= x{x²+(2y)²}{(x²-(2y)²}

= x(x²+4y²)(x+2y)(x-2y) [Answer]

E.4.7 The length of a cubic reservoir is the inverse multiple of the length of another cubic reservoir. The sum of the length of the two reservoirs is 3 feet. What will be the sum of their volume?

Solution:

Let say, the length of 1st cubic resorver = x

According to the question, the length of another cubic reservoir = 1x
And, x + 1   x = 3

or, x² + 1 = 3x [multiplying both sides by x]

or, x² - 3x + 1 = 0

Here, we know,

ax² + bx + c = 0 for that,

x = -b ± √ (b²-4ac)           2a

Therefore, for x, x² - 3x + 1 = 0

x = 3 ± √ {(-3)²-4.1.1}             2.1

or, x = 3 ± √ 5     2

So, x = 0.3819 feet (approx.) or, x = 2.6180 feet (approx.)

So, 1x =      10.3819 = 2.6180 feet (approx.) or, 1x =      12.6180 = 0.3819 feet (approx.)

Therefore, the sum of the volumes of the two cubic reservoir

= x³ + (1x)³

= (0.3819)³ + (2.6180)³

= 0.0556 + 17.9436

= 18 cubic feet (approx.) (Answer)

E.4.8 Express the following fractions in partial fractions. 

i)           x + 1(x - 1)²(x² + 1)² 

ii)  x³ + 1x² + 1

Solution:

i) Please wait, will be posted soon.....

ii)  x³ + 1 x² + 1 

The fraction given here is an improper fraction. Therefore, the fraction can be expressed as a partial fraction using the division method.

x² + 1 ) x³ + 1 ( x
             x³ + x
______________
   (-)      -x + 1

Here, the quotient is x and the numerator is -x + 1

So, x³ + 1x² + 1 

= x + -x + 1x² + 1

= x + -(x - 1) x² + 1

= x -   x - 1x² + 1

That means, x -   x - 1x² + 1 is a partial fraction.