Imagine a situation wherein an engineer is designing a roller coaster. How should he go about it? What part of algebra should he use? He would use polynomials to model the curves for the rollercoaster. Let us take another example wherein a civil engineer has to design a road. What would he do? He too would use polynomials to design roads, buildings and other structures. Polynomials in algebra are an essential tool in our daily lives. Let us know more!

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**Introduction**

You all know, that you can perform four operations on a polynomial that has one variable. You can add, subtract, divide and multiply the polynomials that have over one variable. The process remains just as it is, however, you now have more variable to keep a track on.

So, when you subtract or add a polynomial that has more than one variable, you have to make sure that you combine the like terms only. On the other hand, when you have to divide and multiply, youâ€™ll have to pay specific attention to the multiple terms and variables. You cannot add and subtract, if the terms are not alike. However, in multiplication and division is possible even if the terms arenâ€™t alike.

## How to add polynomials that have more than one variable?

In order to add the polynomials, youâ€™ll first have to identify the like terms in the polynomials. Further, youâ€™ll have to combine these like terms in accordance with the correct integer operations.

Please note that the like terms should have the same exact variables raised to the same exact power, you will have to be extremely careful when identifying a polynomial with more than one variable.

In some cases, parentheses are used so as to differentiate between the addition of 2 polynomials and addition of a variety of monomials. In addition, you can just remove the parentheses, and add.

### Example 1:**Â Â **Add: (4*x*^{2Â }â€“ 12*xy*Â + 9*y*^{2}) + (25*x*^{2Â }+ 4*xy â€“*Â 32*y*^{2})

^{2}

**Step 1: Firstly, youâ€™ll remove the parentheses which group the polynomials and rewrite the subtraction as addition, in case possible. **

i.e. 4*x*^{2Â }+(âˆ’12*xy)*Â + 9*y*^{2}Â + 25*x*^{2Â }+ 4*xy +*Â (âˆ’32*y*^{2})

**Step 2: Group all the like terms using associative and commutative properties. **

i.e. (4*x*^{2Â }+25*x*^{2)}Â +[(âˆ’12*xy*)+ 4*xy*] + [9*y*^{2}*+*Â (âˆ’32*y*^{2})]

**Step 3: Combine all the like terms**

i.e. 29*x*^{2}Â + (âˆ’8*xy*) +(âˆ’23*y*^{2})

**Step 4: Rewrite the balance as subtraction**

Answer: 29*x*^{2}Â â€“Â 8*xyÂ *â€“Â 23*y*^{2}

Some of you might find that writing the polynomial addition in a vertical form makes it easy for you to combine the like terms. In this case, as well, the process of adding the polynomials is same however the only difference is in the arrangement of the terms. Here, in this example below, you can have a look at the â€˜verticalâ€™ method of adding the polynomials.

### Example 2: Add (3x^{Â }+ 2y â€“Â 4zÂ ) + (45x^{Â }â€“Â yÂ + 75z)

Â Â Â 3x + 2y – 4z

+ Â 45x Â – Â y Â + 75z

______________

Â Â Â 48x + y Â + 71z

In this, youâ€™ll have to write one polynomial below the other, making sure that you line up the like terms. Further, add the like terms and youâ€™ll get your answer.**Â **

## How to subtract the polynomials that have more than one variable?

Simply apply the same procedure to subtract the polynomials with one variable to subtract the polynomials that have more than one variable. To remove the parentheses following a subtraction sign, youâ€™ll have to multiply the terms by -1.**Â **

### Example 3: Â Subtract (14*x*^{3}*y*^{2Â }â€“ 5*xyÂ *+ 14*y*) â€“ (7*x*^{3}*y*^{2}Â â€“ 8*xy*Â + 10*y*)

**Firstly: Remove the parentheses and see the signs.**

14x^{3}y^{2Â }â€“ 5xyÂ + 14yÂ â€“ 7x^{3}y^{2}Â + 8xyÂ â€“ 10y

**Secondly: Regroup and arrange the like terms together.**

14x^{3}y^{2Â Â }â€“ 7x^{3}y^{2}Â â€“ 5xyÂ ^{Â }+ 8xyÂ + 14yÂ â€“ 10y

**Thirdly: Combine the like terms**

7x^{3}y^{2}Â + 3xyÂ + 4y

The difference is 7x^{3}y^{2}Â + 3xyÂ + 4y. Again, the alternative to this is the vertical approach. So, here, weâ€™ll solve a problem with a vertical approach to give you an idea of the same.

### Example 4: Subtract (*10a*^{3}Â + 5*b*^{2Â }Â â€“ 5*cÂ *+ 10) â€“ (15 + 5*c*^{Â }Â â€“ 15*b*^{2}Â + 10*a*^{3})

*Â Â 10a*^{3}Â + 5*b*^{2Â }Â â€“ Â 5*c Â *+ 10

– (10*a*^{3 Â }â€“ 15*b*^{2}Â + 5*c*^{Â }Â + 15)

_____________________

Â Â Â Â 0 Â + Â 20*b*^{2 Â }– Â 10c Â – Â 5

## How to multiply polynomials that have more than one variable?

If polynomials have more than one variable, they too can be multiplied by one another. The same technique of multiplying with one variable can be used when you multiply the polynomials with more than one variable.

Letâ€™s take a look at an example of multiplication of two polynomials especially monomials with two variables. In order to do this, youâ€™ll have to multiply the coefficients and use the rules of exponents to find the component for every variable in order to find the product.

For example, (4*x ^{2}y*

^{3})(5

*x*

^{4}

*y*

^{2}) = (4 â€¢ 5)(

*x*

^{2+4})(

*y*

^{3+2}) = 20

*x*

^{6}

*y*

^{5}

Now, when you have to multiply a monomial with a binomial, you will have to use the distributive property in the same manner as multiplying the polynomial with one variable. In order to multiply the two binomials, that have more than one variable, you can still make use of FOIL (First, Outer, Inner, Last) method which works for binomials with one variable.

### Example 5:Â Â Multiply (4*xÂ *â€“ 7*xy*)(2*yÂ *+ 3*x*)

- First: 4
*x*Â â€¢ 2*y*Â = 8*xy* - Outer: 4
*x*Â â€¢ 3*x*Â = 12*x*^{2} - Inner: âˆ’7
*xy*Â â€¢ 2*y =Â*âˆ’14*xy*^{2} - Last: âˆ’7
*xy*Â â€¢ 3*x*Â =Â âˆ’21*x*^{2}*y*

Now, you can combine the terms in one expression 8xyÂ + 12x^{2}Â â€“ 14xy^{2}Â â€“ 21x^{2}y. So, the product is 8xyÂ + 12x^{2}Â â€“ 14xy^{2}Â â€“ 21x^{2}y. Now, letâ€™s take a look at the next example. Here, weâ€™ll have a product of binomial and trinomial each of which will have two variables.

### Example 6: Multiply**Â **(9*b*Â â€“Â *ab*)(5*a*^{2}*b*Â + 7*ab*Â â€“Â *b*)

Solution:Â First: Multiply the term 9b with every term in the bracket

=Â 9*b*(5*a*^{2}*b*Â + 7*ab*Â â€“Â *b*)

= 45*a*^{2}*b*^{2}Â + 63*ab*^{2}Â â€“Â 9*b*^{2}

Secondly: Multiply the term -ab with every term in the bracket

= âˆ’*ab*(5*a*^{2}*b*Â + 7*ab*Â â€“Â *b*)

= âˆ’5*a*^{3}*b*^{2}Â â€“Â 7*a*^{2}*b*^{2}Â + a*b*^{2}

Adding both: 45*a*^{2}*b*^{2}Â + 63*ab*^{2}Â â€“Â 9*b*^{2}Â â€“Â 5*a*^{3}*b*^{2}Â â€“Â 7*a*^{2}*b*^{2}Â + a*b*^{2
}= 45*a*^{2}*b*^{2}Â + 63*ab*^{2}Â â€“Â 9*b*^{2}Â – 5*a*^{3}*b ^{2}*Â â€“Â 7

*a*

^{2}

*b*Â + a

^{2}*b*

^{2 }= 38

*a*

^{2}

*b*

^{2}Â + 64

*ab*

^{2}Â â€“Â 9

*b*

^{2}Â â€“Â 5

*a*

^{3}

*b*

^{2}

So, the product is 38*a*^{2}*b*^{2}Â + 64*ab*^{2}Â â€“Â 9*b*^{2}Â â€“Â 5*a*^{3}*b*^{2}. You can also solve this problem in vertical format.

### Example 7: Solve (9*b*Â â€“Â *ab*)(5*a*^{2}*b*Â + 7*ab*Â â€“Â *b*) in vertical format.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 9b Â – Â ab

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Ã— Â 5a^{2}b Â + Â 7ab Â – Â b

Â Â Â Â Â Â Â Â Â Â _________________________

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â – Â 9b^{2} + Â ab^{2}

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â – Â 7a^{2}b^{2}Â Â Â Â Â Â + 63ab^{2Â }

– 5a^{3}b^{2} + 45a^{2}b^{2}

Â Â Â Â Â Â Â Â Â Â Â _________________________

Â Â Â Â Â Â Â Â Â Â Â Â Â – 5a^{3}b^{2} + 38a^{2}b^{2 }– 9b^{2Â }+ 64ab^{2}

## How to divide polynomials that have more than one variable?

The last operation on polynomial is division. So, if thereâ€™s a polynomial with more than one variable, it can also be divided. So, in here, youâ€™ll have to divide the monomials that have more than one variable and then coefficients and then the variables. If there are exponents with the same base, you will have to divide the exponents by subtracting them according to the law of exponents.

## More Solved Examples for You

**Question 1: Divide the equation given below $$ \frac{14x^3y}{28x^6y^4} $$**

**Answer :** $$ \frac{14}{28} \times Â \frac{x^3}{x^6} \times \frac{y}{y^4} $$

$$ \frac{1}{2} \text(x^{-3}y^{-3})$$

$$ \frac{1}{2x^3y^3} $$

Therefore the quotient will be:

$$ \frac{1}{2x^3y^3} $$

**Question 2: How to solve a polynomial?**

**Answer:** For solving a linear polynomial, first of all, set the equation equal to zero, then isolate and solve for the variable. Moreover, linear polynomial will have only one answer. However, if you have to solve a quadratic polynomial, and then write the equation in order of the highest degree to the lowest, then set the equation equal to zero.

**Question 3: Explain what is polynomial in math?**

**Answer:** It refers to arepresentation that contains variable or indeterminate and coefficients that involve only the operations of addition, multiplication, subtraction, and non-negative integer exponents of variables. For example, a polynomial of single determinate x is 2x^{2} â€“ 5x+ 3.

**Question 4: What is the purpose of learning polynomials?**

**Answer: **We use polynomials to plot complex curves that decide the path of missile trajectories or a roller coaster or model a complex situation in a physics experiment. In addition, its modeling functions can even be used to solve questions in chemistry and biology.

**Question 5: Is pi a polynomial?**

**Answer:** No, pi is not a polynomial because its value refers to the circumference of a circle. Instead, a polynomial refers to an equation that contains four or more variables.

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