Multiplication of a Polynomial by a Monomial
To multiply a polynomial by a monomial, use the distributive
property: multiply each term of
the polynomial by the monomial. This involves multiplying
coefficients and adding exponents of the appropriate variables.
Example 1: 3y^{2}(12y^{3} 6y^{2} + 5y  1) =?
= 3y^{2}(12y^{3}) + (3y^{2})( 6y^{2}) + (3y^{2})(5y) + (3y^{2})( 1)
= (3)(12)y^{2+3} + (3)( 6)y^{2+2} + (3)(5)y^{2+1} + (3)( 1)y^{2}
= 36y^{5} 18y^{4} +15y^{3} 3y^{2}
Example 2: 4x^{3}y( 2y^{2} + xy  x + 9) =?
=  4x^{3}y( 2y^{2}) + ( 4x^{3}y)(xy) + ( 4x^{3}y)( x) + ( 4x^{3}y)(9)
= ( 4)( 2)x^{3}y^{1+2} + ( 4)x^{3+1}y^{1+1} + ( 4)( 1)x^{3+1}y + ( 4)(9)x^{3}y
= 8x^{3}y^{3} 4x^{4}y^{2} +4x^{4}y  36x^{3}y
Multiplication of Binomials
To multiply a binomial by a binomial(a + b)(c + d ), where a,
b, c, and d are termsuse the distributive property twice.
First, treat the second binomial as a single term and distribute over
the first binomial:
(a + b)(c + d )= a(c + d )+ b(c + d ) 

Next, use the distributive property over the second binomial:
a(c + d )+ b(c + d )= ac + ad + bc + bd 

At this point, there should be
4 terms in the answer  every
combination of a term of the first binomial and a term of the second
binomial. Simplify the answer by combining like terms.
We can use the word FOIL to remember how to multiply two binomials (a + b)(c + d ):
 Multiply their First terms. (ac)
 Multiply their Outside terms. (ad )
 Multiply their Iinside terms. (bc)
 Multiply their Last terms. (bd )
 Finally, add the results together: ac + ad + bc + bd. Combine like terms.
Remember to include negative signs as part of their respective terms in the multiplication.
Example 1.(xy + 6)(x + 2y) =?
= (xy)(x) + (xy)(2y) + (6)(x) + (6)(2y)
= x^{2}y + 2xy^{2} + 6x + 12y
Example 2.(3x^{2} +7)(4  x^{2}) =?
= (3x^{2})(4) + (3x^{2})( x^{2}) + (7)(4) + (7)( x^{2})
= 12x^{2} 3x^{4} +28  7x^{2}
=  3x^{4} + (12  7)x^{2} + 28
=  3x^{4} +5x^{2} + 28
Example 3: (y  x)( 4y  3x) =?
= (y)( 4y) + (y)( 3x) + ( x)( 4y) + ( x)( 3x)
=  4y^{2} 3xy + 4xy + 3x^{2}
= 3x^{2} + ( 3 + 4)xy  4y^{2}
= 3x^{2} + xy  4y^{2}
Multiplication of Polynomials
The strategy for multiplying two polynomials in general is similar to
multiplying two binomials. First, treat the second polynomial as a
single term, and distribute
over the first term:
(a + b + c)(d + e + f )= a(d + e + f )+ b(d + e + f )+ c(d + e + f ) 

Next, distribute over the second polynomial:
a(d + e + f )+ b(d + e + f )+ c(d + e + f )= ad + ae + af + bd + be + bf + cd + ce + cf 

At this point, the number of terms in the answer should be the number
in the first polynomial times the number in the second polynomialevery
combination of a term of the first polynomial and a term of the
second polynomial. Since there are
3 terms in each polynomial in this
example there should be
3(3) = 9 terms in our answer so far. If the
first polynomial had
4 terms and the second had
5, there would be
4(5) = 20 terms in the answer so far.
Finally, since the the terms in such a product of polynomials are often
highly redundant (many have the same variables and exponents), it is important
to combine like terms.
Example 1: (x^{2} 2)(3x^{2}  3x + 7) =?
= x^{2}(3x^{2} 3x + 7)  2(3x^{2}  3x + 7)
= x^{2}(3x^{2}) + x^{2}( 3x) + x^{2}(7)  2(3x^{2})  2( 3x)  2(7) (6 terms)
= 3x^{4} 3x^{3} +7x^{2} 6x^{2} + 6x  14
= 3x^{4} 3x^{3} + (7  6)x^{2} + 6x  14
= 3x^{4} 3x^{3} + x^{2} + 6x  14
Example 2: (x^{2} + x + 3)(2x^{2}  3x + 1) =?
= x^{2}(2x^{2} 3x + 1) + x(2x^{2} 3x + 1) + 3(2x^{2}  3x + 1)
= x^{2}(2x^{2}) + x^{2}( 3x) + x^{2}(1) + x(2x^{2}) + x( 3x) + x(1) + 3(2x^{2}) + 3( 3x) + 3(1) (9 terms)
= 2x^{4} 3x^{3} + x^{2} +2x^{3} 3x^{2} + x + 6x^{2}  9x + 3
= 2x^{4} + ( 3 + 2)x^{3} + (1  3 + 6)x^{2} + (1  9)x + 3
= 2x^{4}  x^{3} +4x^{2}  8x + 3
Note: To check your answer, pick a value for the variable and
evaluate both the original expression and your answerthey should
be the same.