


LESSON 4:
OPERATIONS WITH POLYNOMIALS
Addition and Subtraction:
Example
1: “+” sign preceding brackets – simply drop the brackets and collect like terms
a) (3x^{2} – 2x + 5) + (5x^{2} –
3x – 6) = 3x^{2} – 2x + 5 + 5x^{2} – 3x – 6 ** drop
brackets
= 3x^{2} + 5x^{2} –2x – 3x
+5 – 6 ** Collect
like terms
= 8x^{2 }– 5x – 1
Example
2: “” sign preceding brackets – multiply each term in the bracket by “1” and collect like terms
(5x^{2}
– 3x + 6) – (2x^{2} – 7x + 8) = (5x^{2} – 3x + 6) – 1(2x^{2}
– 7x + 8) ** multiply 2^{nd}
bracket by “1”
=
5x^{2} – 3x + 6 – 2x^{2} + 7x  8
=
5x^{2} – 2x^{2} – 3x + 7x + 6 – 8
** collect like terms
=
3x^{2} + 4x  2
Example
3:
a) (7a^{2 }– 3ab – 5b^{2}) + ( 4a^{2} – 8ab – 2b^{2}) = 7a^{2 }–
3ab – 5b^{2} + 4a^{2} – 8ab – 2b^{2}
= 7a^{2} + 4a^{2} – 3ab –
8ab – 5b^{2} – 2b^{2}
= 11a^{2} – 11ab – 7b^{2}
b) (8x + 3y
– 5xy) – (4x – 7y – 6xy) = (8x + 3y – 5xy) – 1(4x – 7y – 6xy)
** multiply 2^{nd} bracket by “1”
= 8x + 3y – 5xy – 4x + 7y + 6xy
= 8x – 4x + 3y + 7y – 5xy + 6xy ** collect
like terms
= 4x + 10y + xy
Multiplying with Polynomials
(Expanding):
Example
1: Monomial x Polynomial – multiply each term in bracket by the monomial
a) –3x(2x^{2} – 5x + 7) = 3x(2x^{2}) – 3x(5x)
– 3x(7) ** multiply each term by “3x”
= 6x^{3 }+ 15x^{2} –
21x
b) 2x(5x – 3) –5(2x +
7) = 10x^{2} – 6x – 10x – 35 ** multiply each term in 1^{st} bracket by “2x” and 2^{nd}
bracket by “5”
= 10x^{2} – 16x – 35 ** collect
like terms
Example
2: Polynomial x Polynomial – multiply each term in 1^{st} bracket by each term in 2^{nd} bracket
a) (3x + 5)(2x – 7) = 3x(2x – 7) + 5(2x – 7) ** multiply each term in 1^{st} bracket by each term in 2^{nd} bracket
= 6x^{2} – 21x + 10x –35 ** expand as in previous example
b) (2x + 3)(3x^{2} – 5x – 2) = 2x(3x^{2} – 5x – 2) +
3(3x^{2} – 5x – 2) ** multiply each term in 1^{st} bracket by each term in 2^{nd} bracket
= 6x^{3} –10x^{2} – 4x + 9x^{2}
–15x – 6 ** expand
= 6x^{3} – x^{2} –19x – 6 ** collect like terms
c) 2(5x –
3)^{2} – 3(2x – 1)(3x + 2) = 2(5x – 3)(5x – 3)
– 3(2x – 1)(3x + 2)
= 2(25x^{2} – 15x – 15x + 9) – 3(6x^{2}
+ 4x – 3x – 2)
= 50x^{2} – 30x – 30x + 18 – 18x^{2}
– 12x + 9x + 6
= 32x^{2} – 63x + 24
Factoring Polynomials
Review of Basic Factoring methods:
1.
Common Factoring:
Factor 6x^{3}
– 15x
Solution:
6x^{3} – 15x = 3x(2x^{2}
– 5)
** Find the HIGHEST COMMON FACTOR for each term  “3x”
** Divide “3x” into each term to get the second factor  “2x^{2} – 5”
** Check by expanding
2.
Difference of Squares: Formula  a^{2}
– b^{2 }= (a – b)(a + b)
Factor 49x^{2}
– 64y^{2}
Solution:
49x^{2} – 64y^{2} = (7x – 8y)(7x + 8y)
Factor x^{2}
– 9y^{2}
Solution:
x^{2} – 9y^{2}
= (x – 3y)(x + 3y)
3.
Simple Trinomials: Form x^{2} + bx +
c [Coefficient of x^{2 }is 1]
a) Factor x^{2}
+ 5x + 6
Solution:
Recall x^{2} + 5x + 6 = (x + __ )(x + __ ) ** We
need two numbers that multiply to +6 and add to +5
** Check all pairs of factors of 6: {1,
6} adds to 7
{2, 3}
adds to 5
Hence x^{2}
+ 5x + 6 = (x + 3 )(x + 2) ** Check by expanding
b) Factor x^{2}
– 2x – 35
Solution:
Recall x^{2}  2x  35 = (x + __ )(x + __ ) ** We
need two numbers that multiply to 35 and add to 2
** Check all pairs of factors of 35: {1,
35} no
combination adds to “2”
{5, 7} 7 + 5 = 2
Hence x^{2}
– 2x  35 = (x + 5 )(x – 7) ** Check by expanding
4.
Hard Trinomials: Form ax^{2} + bx + c [Coefficient of x^{2 } does not equal 1]
a) Factor
6m^{2} – 5m – 4
Solution:
We use the
method of decomposition (although there are other methods)
We
decompose the middle term “5m” into two parts using the two clues:
Multiply to (6)(4)
= 24 and
Add to “5”
Strategy:
List all pairs of
factors of “24” and find the pair that adds to “5” ** {1, 24} cannot obtain “5” with these two
factors; 1+24=25; 124= 23
** {2,
12} cannot obtain “5” with these two factors
** {3,
8} 8 + 3 = 5 Choose these two factors
Hence 6m^{2} – 5m – 4 = 6m^{2} – 8m + 3m
–4 ** “5m” broken into two parts “8m + 3m”
= 2m(3m – 4) + 1(3m – 4) ** Group by twos and common factor
= (3m – 4)(2m + 1) ** check by expanding
b) Factor
2x^{2} – 9x + 4
Solution:
We
decompose the middle term “9x” into two parts using the two clues:
Multiply to (2)(4)
= +8 and
Add
to “9”
Strategy:
List all pairs of
factors of “+8” and find the pair that adds to “9” ** {2, 4} cannot obtain “9” with these two
factors; 2+4=6; 24= 2
** {1, 8}
8 + 1 = 9 Choose these
two factors
Hence 2x^{2} – 9x + 4 = 2x^{2} –8x – 1x
+ 4 ** “9x” broken into two parts “8x – 1x”
= 2x(x – 4)
– 1(x – 4) ** Group by twos and common factor
= (x – 4)(2x – 1) ** check by expanding