Polynomial multiplication is a very common operation throughout Algebra and Mathematics in general. We use following three properties very frequently all the way along as we work on multiplication of polynomials.
- Associative Properties of Algebra
- Distributive Properties of Algebrat
- Commutative Properties of Algebra
- Rules / Laws of Exponents
We have already discussed the first three properties in the preceding topics.
We enlist the relevant formulas below so that you can revise these properties quickly:
Commutative Law For Addition
`a + b = b + a`
Commutative Law For Multiplication
`a*b = b * a`
Associative Law For Addition
`a + ( b + c ) = ( a + b ) + c`
Associative Law For Multiplication
`a * ( b *c ) = ( a * b ) * c`
`a *( b + c ) = ( a* b ) + ( a * c )`
What remains is the rules of exponents. We explain these rules first and then we move forward to polynomial multiplication.
Let a be a real number. And let m,n be any positive integers. Then:
`a^m * a^n = a^m+n `
If a is any real number, and m,n are two positive integers:
` (a^m)n = a^mn `
Let a,b be two real numbers. And let n be a positive integer then:
` (ab)^n = a^n.b^n`
These properties are very simple and easy to verify. You must learn them by heart before you move ahead towards the multiplication of polynomials.
`x^2x^7 = x^(2+7) = x^9`
`7x^3` . `5xy^2 = (7)(5)x^(3+1)y^2 = 35x^4y^2`
`x^4x^2x^3x = x^(4+2+3+1) = x^10`
`2y^3x^2`.`7xy^2 = (2)(7)(x^(2+1))(y^(3+2)) = 14x^3y^5`
`(-3yz^2)(7x^3y^4z) = (-3)(7)x^3 (y^(1+4))(z^(1+2 )) = -21x^3y^5z^3`
`(x^2y^2)^3 = (x^2)^3 (y^2)^3 = x^(2.3) y^(2.3) = x^6y^6 `
`(2yz^2)^4 = (2)^4 y^4 (z^2)^4 = 16y^4 z^(2.4) = 16y^4z^8`
These are some of the most basic rules for polynomial multiplication. Now that you have practiced these simplifications, the process of multiplication is going to be very easy for you
The simple rule that we use while multiplying two polynomials is that
While multiplying these terms, we make use of the rules of exponents stated above, whereas the remaining formulas help us simplify and / or expand the multiplying polynomials.
We start with the simpler examples (involving monomials) and then proceed towards more complex examples (those involving polynomials with 2 and more terms).
We have solved this type of examples in the previous section.
So this multiplication is straight-forward.
=> `(4x^2)( -3x^4) = (4)( -3)(x^(2+4)) = -12x^6`
In this type of multiplication, we make use of the Distributive Property of Addition.
=> `(2x)(4x^2 + 3xy)`= `(2x)(4x^2) + (2x)(3xy)` = `(2)(4)(x.x^2) + (2)(3)(x.xy)` = ` 8x^(1+2) + 6x^(1+1)y` = `8x^3 + 6x^2y`
In order to find this product, we multiply the term outside parentheses `( -x^2 )` by each term inside the parentheses `( 5x^2 + 2x - 2 )` , and we also take care of various signs involved during multiplication.
But wait. What is the coefficient of `-x^2` ? It is basically 1. This means that `-x^2 = -1x^2`
In order to emphasize this point, we place 1 with each multiplying term below:
=> ` - 1x^2 ( 5x^2 + 2x - 2 )``(-1x^2 )( 5x^2 ) + (-1x^2 )( 2x ) + (-1x^2 )( -2 )` = ` (-1)( 5 ) x^2.x^2 + (-1)( 2 ) x^2.x + (-1)( -2 ) x^2 ` = ` - 5x^4 - 2x^3 + 2x^2 `
In this case, we are supposed to multiply a trinomial (a polynomial with three terms) with a binomial (a polynomial with two terms). For this type of multiplication, we implement the rule of "Multiply each term in the first polynomial by each term in the second polynomial".
=> ` ( x^3 + 6x - 1 ) ( 2x^2 + 4 ) = x^3 ( 2x^2 + 4 ) + 6x( 2x^2 + 4 ) `
`-1( 2x^2 + 4 )`
Now we use the Distributive property on each of these three terms:
Now we simplify these terms through rules of Exponents:
=> `( x^3 + 6x - 1 ) ( 2x^2 + 4 ) = (1)(2)(x^3.x^2) + (1)(4)x^3 + (6)(2)x.x^2`
`+ (6)(4)x + (-1)(2)x^2 - 4`
=>` ( x^3 + 6x - 1 ) ( 2x^2 + 4 ) = 2x^5 + 4x^3 + 12x^3 + 24x - 2x^2 - 4`
We reorder the terms in descending powers of x and also combine the like terms:
=> ` ( x^3 + 6x - 1 ) ( 2x^2 + 4 ) = 2x5 + ( 4 + 12 )x^3 - 2x^2 + 24x - 4`
=> ` ( x^3 + 6x - 1 ) ( 2x^2 + 4 ) = 2x^5 + 16x^3 - 2x^2 + 24x - 4`
We multiply two binomials in the same fashion as we multiplied a binomial with a trinomial. Using this method:
`(3x + 4)(2x - 1)`= `3x(2x - 1) + 4(2x - 1)` = ` 3x(2x) + 3x(-1) + 4(2x) + 4(-1) ` = `(3)(2)x.x - 3x + 8x - 4` = `6x^2 + 5x - 4`
FOIL Method is a reserved word for a method of multiplying two binomials. The four letters of the word FOIL each refer to the following four operations:
|FIRST||Multiply together the "First" terms of each of the two binomials.|
|OUTER||Multiply together the "Outermost" terms of each of the two polynomials.|
|INNER||Multiply the "Inner" terms - that is the second term of the first binomial and the first|
|LAST||Multiply the "last" terms of each of the two binomials.|
All the resultant terms are then combined and the like terms are simplified to obtain the final result.
Generally, the formula for FOIL method is given as follows:
We can prove the foil method using the multiplication method we have studied above. Let us demonstrate this proof here:
` (a + b)(c + d) ` = ` a(c + d) + b(c + d) ` = `ac + ad + bc + bd`
Which is the same result as the one shown above.
We solve a few examples of binomial multiplication using FOIL method:
We use FOIL method to find the product:
1. Multiply First terms: `x.x = x^2 `
2. Multiply Outer terms: `x.5 = 5x`
3. Multiply Inner terms: `(-3)(x) = - 3x`
4. Multiply Last terms: `(-3)(5) = -15`
Next we need to add all these terms together and combine like terms to get the final product:
=> ` (x - 3)(x + 5)`= ` x^2 + 5x - 3x -15` = ` x^2 + 2x -15`
Using the FOIL method of binomial multiplication
1. Multiply First terms: `(2x)(3x) = (2)(3)x.x = 6x^2`
2. Multiply Outer terms: `(2x)(1) = 2x`
3. Multiply Inner terms: `(- 5)(3x) = (-5)(3)x = -15x`
4. Multiply Last terms: `(-5)(1) = - 5`
Next we add these terms and combine the like terms as we did before:
=> `(2x - 5)(3x + 1)`= ` 6x^2 + 2x + (-15x) + (- 5)` = ` 6x^2 + 2x - 15x - 5` = `6x^2 - 13x - 5`
Till this point, we have seen only the horizontal method of polynomial multiplication. Just like we saw vertical method of addition and subtraction, we can multiply polynomials through vertical method as well.
In order to understand the vertical method, you are first supposed to recall the elementary multiplication method. Let us present an example underneath:
1 1 4
2 2 8
1 1 4
1 3 6 8
That is, we stack the two numbers above and below each other. And then we take one digit from the lower number (starting from the right towards left) and multiply it through each digit in the upper number (from right to left). For each new digit in the lower number, we form a new row and each new row is shifted towards left by one place.
After we are done with all the digits in the lower number, the rows obtained are added together to produce the required result of multiplication.
In vertical method of polynomial multiplication, we follow the same procedure. The only difference is that we multiply "terms" instead of digits.
x - 3
x + 5
5x - 15
` x^2 - 3x `
` x^2 + 2x - 15`
Hence we have obtained the same result through vertical method as the one we obtained using horizontal method of polynomial multiplication. However, vertical method is really helpful when you are supposed to multiply larger-sized polynomials.
We present an example next.
We solve using Horizontal method first:
`(x - 4)(3x + y - 2)`= ` x (3x + y - 2) - 4(3x + y - 2)` = ` 3x^2 + xy - 2x - 4(3x) - 4y - 4(- 2)` = `3x^2 + xy - 2x - 12x - 4y + 8` = ` 3x^2 + xy - 14x - 4y + 8`
which is the desired product of the given polynomials. Now we solve the same example using vertical method:
` 3x + y - 2`
` x - 4`
` - 12x - 4y + 8 ` (Multiplication with - 4)
` 3x^2 + xy - 2x ` (Multiplication with x)
` 3x^2 + xy - 14x - 4y + 8`
Which is the same result as before. But vertical method makes multiplication remarkably easy.
Note that we have placed the like terms above and below each other in order to facilitate their addition.
In the given problem, both multiplying polynomials have 3 terms each. When we need to multiply such larger sized polynomials, it is always preferable to choose the vertical method. You will notice shortly that vertical method makes this problem very easy to solve.
`x^3 + x^2 + 2`
` 2x^3 + 3x + 4`
`4x^3 + 4x^2 + 8`
` 3x^4 + 3x^3 + 6x `
` 2x^6 + 2x^5 + 4x^3`
` 2x^6 + 2x^5 + 3x^4 + 11x^3 + 4x^2 + 6x + 8 `
Hence we get the polynomial ` 2x^6 + 2x^5 + 3x^4 + 11x^3 + 4x^2 + 6x + 8` through the multiplication of given polynomials.
You can try and solve the same problem using horizontal method as well, but doing so will further emphasize the ease of vertical method.