Category: Mathematics

Monomial is an algebra expression that contains one term. It includes numbers, whole numbers, variables, and all are multiplied together. The highest common factor (HCF) or the greatest common factor (GCF) of two monomials are the product of numerical coefficients and literal coefficients.

Steps to Find HCF of Monomials

The following are the simple steps to calculate the greatest common factor of two monomials easily.

• Get the factors of both monomial numerical coefficients.
• From that, find the HCF of the numerical coefficient.
• Coming to the literal coefficients, check the lowest power of each variable.
• And calculate the HCF of literal coefficients.
• Multiply the HCF of numerical coefficients and literal coefficients.

Example Questions on Highest Common Factor of Monomials

Example 1.

Find the H.C.F. of 25x³y² and 5xy³z.

Solution:

The H.C.F. of numerical coefficients = The H.C.F. of 25 and 5.

Since, 25 = 5 × 5 = 5² and 5 = 1 × 5 = 5¹

Therefore, the H.C.F. of 25 and 5 is 5

The H.C.F. of literal coefficients = The H.C.F. of x³y² and xy³z = xy²

Since, in x³y² and xy³z, x and y are common.

The lowest power of x is x.

The lowest power of y is y².

Therefore, the H.C.F. of x³y² and xy³z is xy².

Thus, the H.C.F. of 25x³y² and 5xy³z.

= The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 5 × (xy²)

= 5xy².

Example 2.

Find the G.C.F of 32x²yz² and 72xy²z

Solution:

The H.C.F. of numerical coefficients = The H.C.F. of 32 and 72.

Since, 32 = 2 x 2 x 2 x 2 x 2 = 2⁵ and 72 = 2 x 2 x 2 x 9 = 2³ x 9

Therefore, the H.C.F. of 32 and 72 is 2³ = 8

The H.C.F. of literal coefficients = The H.C.F. of x²yz² and xy²z = xyz

Since, in x²yz² and xy²z, x, y and z are common.

The lowest power of x is x.

The lowest power of y is y.

The lowest power of z is z.

Therefore, the H.C.F. of x³y² and xy³z is xyz.

Thus, the H.C.F. of 32x²yz² and 72xy²z.

= The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 8 × (xyz)

= 8xyz.

Example 3.

Find the highest common factor of 8p²q⁵r³ and 6p⁴q³r².

Solution:

The H.C.F. of numerical coefficients = The H.C.F. of 8 and 6.

Since, 8 = 2 x 2 x 2 = 2³ and 6 = 2 × 3 = 2¹ x 3¹

Therefore, the H.C.F. of 8 and 6 is 2

The H.C.F. of literal coefficients = The H.C.F. of p²q⁵r³ and p⁴q³r² = p²q³r²

Since, in p²q⁵r³ and p⁴q³r², p, q, and r are common.

The lowest power of p is p².

The lowest power of q is q³.

The lowest power of r is r²

Therefore, the H.C.F. of p²q⁵r³ and p⁴q³r² is p²q³r².

Thus, the H.C.F. of 8p²q⁵r³ and 6p⁴q³r².

= The H.C.F. of numerical coefficients × The H.C.F. of literal coefficients

= 2 × (p²q³r²)

= 2p²q³r².

Do you need any help to find the least common multiple of polynomials? Then read this article to find the L.C.M of polynomials by factorization method. All you need to do is get the factors of each polynomial, multiply the common and remaining terms to get the L.C.M. You can also check the example questions in the below sections.

LCM of Polynomials Solved Examples

Example 1.

Find the lowest common multiple of (2x² – 4x), (3x⁴ – 12x²), and (2x⁵ – 2x⁴ – 4x³)?

Solution:

Given polynomials are (2x² – 4x), (3x⁴ – 12x²), and (2x⁵ – 2x⁴ – 4x³).

First Polynomial = (2x² – 4x)

= 2x(x – 2), by taking 2x common

Second Polynomial = (3x⁴ – 12x²)

= 3x² (x² – 4), by taking 2x² common.

= 3x²(x² – 2²), by using the formula a² – b² = (a + b) (a – b)

= 3x²(x + 2) (x – 2)

Third Polynomial = (2x⁵ – 2x⁴ – 4x³)

= 2x³(x² – x -2), by taking 2x³ common

= 2x³(x² – 2x + x – 2), by splitting the middle term -x = -2x + x

= 2x³(x(x – 2) + 1(x – 2))

= 2x³(x + 1) ( x – 2)

The common terms of (2x² – 4x), (3x⁴ – 12x²), and (2x⁵ – 2x⁴ – 4x³) = x(x – 2)

Extra common terms are 2, 3x(x + 2), 2x²(x + 1)

Therefore, the required L.C.M. = x(x – 2) * 2 * 3x(x + 2) * 2x²(x + 1)

= 12x⁴ (x – 2) (x + 2) (x + 1).

Example 2.

Find the L.C.M of 3y³ – 18y²x + 27yx², 4y⁴ + 24y³x + 36y²x² and 6y⁴- 54y²x² by factorization.

Solution:

Given polynomials are 3y³ – 18y²x + 27yx², 4y⁴ + 24y³x + 36y²x² and 6y⁴- 54y²x²

First polynomial = 3y³ – 18y²x + 27yx²

= 3y(y² – 6yx +9x²) by taking 3y common

= 3y(y² – 3yx -3yx +9x²) by splitting the middle term -6yx = -3yx – 3yx

= 3y(y(y – 3x) -3x(y – 3x))

= 3y(y – 3x) (y – 3x)

Second Polynomial = 4y⁴ + 24y³x + 36y²x²

= 4y²(y² + 6yx + 9x²), by taking 4y² common

= 4y²(y² + 3yx + 3yx + 9x²) by splitting the middle term 6yx = 3yx + 3yx

= 4y²(y(y + 3x) + 3y(y + 3x))

= 4y²(y + 3x) (y + 3x)

Third Polynomial = 6y⁴- 54y²x²

= 6y²(y² – 9x²), by taking 6y² common

= 6y²(y² – (3x)²) by using a² – b² formula

= 6y²(y + 3x) (y – 3x)

The common factors of the above three expressions is ‘y’ and other common factors of first and third expressions are ‘3’ and ‘(y – 3x)’. The common factors of second and third expressions are ‘2’, ‘y’ and ‘(y + 3x)’. Other than these, the extra common factors in the first expression is ‘(y – 3x)’ and in the second expression are ‘2’ and ‘(y + 3x)’

Therefore, the required L.C.M. = y × 3 × (y – 3x) × 2 × y × (y + 3x) × (y – 3x) × 2 × (y + 3x)

= 12y²(y + 3x)² (y – 3x)²

Example 3.

Claculate the L.C.M of 2x² -x -1 and 4x² + 8x +3?

Solution:

Given polynomials are 2x² -x -1 and 4x² + 8x +3.

First Polynomial = 2x² -x -1

= 2x² – 2x + x -1, by splitting the middle term -x = -2x + x

= 2x(x – 1) + 1 (x – 1)

= (x – 1) (2x + 1)

Second Polynomial = 4x² + 8x +3

= 4x² + 6x + 2x + 3, by splitting the middle term 8x = 6x + 2x

= 2x(2x + 3) + 1(2x + 3)

= (2x + 1)(2x + 3)

The common factors of the above two expressions is (2x + 1). The extra common factor are (2x + 3), (x – 1).

Therefore, the required L.C.M. = (2x + 1) * (x – 1) * (2x + 3)

= (2x + 1) (x – 1) (2x + 3)

No matter how many polynomials are given, you can easily find the great common factor of those polynomials by finding the factors. Find the factors of numerical and literal coefficients. Multiply the common factors out of all polynomials to get the highest common factor. Check out some example questions and answers on HCF of Polynomials by Factorization.

Solved Examples on G.C.F of Polynomials

Example 1.

Find the Greatest Common Factor of x² – 2x + 2, x⁴ – 1, x³ – 2x² – 5x + 6?

Solution:

Factorizing x² – 2x + 2 by using (a – b)².

= (x – 1)²

= (x – 1) ( x – 1)

Factorizing x⁴ – 1 by using a² – b².

= (x²)² – 1²

= (x² – 1) (x² + 1)

= ((x)² – 1²) (x² + 1)

= (x + 1) (x – 1) (x² + 1)

Factorizing x³ – 2x² – 5x + 6 by splitting the middle terms

= x³ – x² – x² – 6x + x + 6

= x²(x – 1) – x (x – 1) – 6 (x – 1)

= (x – 1) (x² – x – 6)

= (x – 1) (x² – 3x + 2x – 6)

= (x – 1) (x(x – 3) + 2 (x – 3))

= (x – 1) (x – 3) (x + 2)

The common factor of x² – 2x + 2, x⁴ – 1, x³ – 2x² – 5x + 6 is (x – 1)

Therefore, H.C.F of x² – 2x + 2, x⁴ – 1, x³ – 2x² – 5x + 6 is (x – 1).

Example 2.

Calculate the Highest Common Factor of x²y² – x² and xy² – 2xy – 3x by factorization.

Solution:

Factorizing x²y² – x² by taking the x² common.

= x² (y²- 1)

= x² (y² – 1²)

= x² ( y + 1) ( y – 1)

= x * x ( y + 1) ( y – 1)

Factorizing xy² – 2xy – 3x by taking the variable x common.

= x (y² – 2y – 3)

= x (y² – 3y + y – 3)

= x (y(y – 3) + 1( y – 3))

= x (y – 3) ( y + 1)

The common factors of x²y² – x² and xy² – 2xy – 3x are x , (y – 1).

Therefore, the H.C.F of x²y² – x² and xy² – 2xy – 3x is x * (y – 1).

Example 3.

Find the H.C.F of x⁴ – y⁴ and x²(x – y) + y²(x – y) + y – x.

Solution:

Factorizing x⁴ – y⁴ by using a² – b² formula.

= (x²)² – (y²)²

= (x² + y²) (x² – y²)

= (x² + y²) (x + y) ( x – y)

Factorizing x²(x – y) + y²(x – y) + y – x by taking (x – y) common

= (x – y) (x² + y² – 1)

The common factor of x⁴ – y⁴ and x²(x – y) + y²(x – y) + y – x are (x – y)

Therefore, the H.C.F of x⁴ – y⁴ and x²(x – y) + y²(x – y) + y – x is (x – y).

Let us learn in detail how to calculate compound interest with the growing principal in this article. If the Interest is due at the end of a certain period be it after a year, half-year, quarterly duration, etc. to a moneylender then Interest is added to the sum borrowed. Thus, Amount becomes the Principal for the next period of borrowing. The Procedure continues further till the specific time is found.

Refer to the Solved Examples on finding Compound Interest with Growing Principal and understand the concept better. We have provided step by step solutions for all the Compound Interest with Increasing Principal Problems.

Solved Examples on Compound Interest with Growing Principal

1. A man takes a loan of $ 20,000 at a compound interest rate of 5% per annum.

(i) Find the amount after 1 year?

(ii) Find the compound interest for 2 years?

(iii) Find the amount required to clear the debt at the end of 2 years?

Solution:

(i) Given P = $20, 000, R = 5%, T = 1 year

We know the Formula for Amount A = P(1+R/100)ⁿ

Substitute the given input data in the formula to obtain the amount

A = 20000(1+5/100)¹

= 20000(1+0.05)

= 20000(1.05)

= 21000

Amount to be paid after 1 year is $21, 000.

SI = A – P

= 21, 000 – 20, 000

= $1, 000

(ii) For Second Year New Principal = $21,000

Thus, Interest in Second Year = 5% of 21, 000

= 5/100*21, 000

= $1050

Compound Interest for 2 Years = Interest of 1st Year + Interest of 2nd Year

= $1000+$1050

= $2050

(iii) Amount required to clear the debt at the end of 2 years = Principal + Compound Interest for 2 Years

= $20, 000 + $2050

= $22, 050

2. At 8% per annum, find the compound interest for 2 years on a certain sum of money?

Solution:

Let the sum be x

Interest for 1st year = 8% of x

= 8x/100

Amount after 1 year = Principal + Interest after 1 year

= x+8x/100

= 108x/100

For second-year New Principal is 108x/100

Interest for 2nd Year = 8% of 108x/100

= 8/100*108x/100

=864x /10000

Amount after 2 years = Principal + Interest after 2 Years

= x +864x/100

= 964x/100

CI = Interest for 1st year + Interest for 2nd Year

=8x/100 + 864x /10000

= 8x/100+54x/625

= (200x+216x)/2500

= 416x/2500

= 104x/625

3. Calculate the compound interest to be paid on a loan of Rs. 5000 for 5/2 years at 10% per annum compounded half-yearly?

Solution:

From the given data Principal = Rs. 5000

T = 5/2

R = 10%

A = P(1+R/100)ⁿ

When Interest is Compounded Half Yearly we need to multiply T with 2 and divide R with 2 thus equation becomes as such

A = 5000(1+10/2*100)^2*5/2

= 5000(1+5/100)⁵

= 5000(105/100)⁵

= 5000(1.2762)

= 6381

CI = A – P

= 6381 – 5000

= Rs. 1381

Therefore, Compound Interest on a sum Rs. 5000 for 5/2 years at 10%, when compounded half-yearly, is Rs. 1381.

Compound Interest, in general, is the Interest Calculated on a Principal and the Interest Accumulated over the Previous Period. It is not similar to the Simple Interest where Interest is not added. Get to know the Compound Interest Formula, Procedure on How to find the Compound Interest on a Daily, Monthly, Quarterly, Yearly Basis. Check out the Solved Examples explained step by step for a better understanding of the concept.

Compound Interest Definition

In Simple Words, Compound Interest is nothing but the Interest that adds back to the Principal Sum so that Interest will be earned during the next compounding period.

Formula to Calculate Compound Interest

Compound Interest Formula is given by Compound Interest = Amount – Principal

Amount A = P(1+r/n)^nt

where,

A= Amount

P= Principal

R= Rate of Interest

n = number of times the interest is compounded per year.

Compound Interest When Interest is Compounded Annually

The Amount Formula mentioned above is the general formula for the number of times the principal is compounded in a year. If the Amount is Compounded Yearly or Annually then Amount Formula is given by

A= P(1+R/100)^t

Compound Interest (CI) when Interest is Compounded Half-Yearly

In this case, if  R is the Rate of Interest Per Annum then it is clearly R/2 per half-year.
A = P(1 + (R/2)/100)^2*1

R/2 = R^‘

CI = A – P

= P(1 + (R/2)/100)^2*1 – P

In the Cases, When the Rate is compounded Half-Yearly, we divide the rate by 2 and multiply the time by 2 before using the general formula for the amount in case of the compound interest.

Compound Interest when Interest is Compounded Quarterly

Let us consider the Compound Interest on a Principal P kept for 1 Year and Interest Rate is R %. Since the Interest is Compounded Quarterly Principal Amount will be changed after 3 months. Interest for the Next Three Months will be calculated on the Amount after 3 first months.

In the same way, Interest for Third Quarter will be calculated on the amount left after the first 6 months. Last Quarter will be calculated on the amount remaining after the first 9 months.

A = P(1+ (R/4)/100)^4T

CI = A – P

= P(1+ (R/4)/100)^4T – P

Solved Example Questions on Compound Interest

1. A town had 15,000 residents in 2000. Its population declines at a rate of 10% per annum. What will be its total population in 2004?

Solution:

The population of a town decreases by 10% every year. Thus, the population of a town next year is calculated on the current year. For Decrease, we have the formula

A = P(1-R/100)ⁿ

= 15000(1-10/100)⁴

= 15000(0.9)⁴
= 9841

The population of the town in 2004 is 9841.

2. The price of a radio is Rs 2000 and it depreciates by 5% per month. Find its value after 4 months?

Solution:

For depreciation, the Amount is A = P(1-R/100)ⁿ

= 2000(1-5/100)⁴

= 2000(1-0.05)⁴

=2000(0.95)⁴
= 1629

Price of radio is Rs. 1629 after 4 months.

3. Calculate the compound interest (CI) on Rs. 10000 for 2 years at 5% per annum compounded annually?

Solution:

We know the formula for Compound Interest Annually is

A= P(1+R/100)^t

= 10000(1+5/100)²

= 10000(105/100)²

= 10000(1.1025)

= 11025

CI = A – P

= 11025 – 10000

= Rs. 1025

4. Calculate the compound interest to be paid on a loan of Rs. 5000 for 3/2 years at 10% per annum compounded half-yearly?

Solution:

Rate of Interest when compounded half-yearly we need to divide R by 2 and multiply Time with 2.

A = P(1+R/100)ⁿ

= P(1+R/2*100)^2*n

= 5000(1+10/2*100)^2*3/2

= 5000(1+5/100)³

= 5000(105/100)³

= 5000(1.157)

= 5788

CI = A – P

= 5788 – 5000

= 788

The Compounded Interest to be paid on a loan is Rs.788