Classify the following polynomials as binomials, trinomials. Which polynomials do not fit in any of these three categories? x + y, 1000, x + x2 + x3 + x4 ,7 + y + 5x, 2y – 3y2, 2y – 3y2 + 4y3, 5x – 4y + 3xy, 4z – 15z2, ab + ac + cd + da, pqr, p2q + pq2, 2p + 2q
Answers
Answer: 1000 and pqr
Step-by-step explanation:
Binomials: x+y, 2y-3y2, p2q+pq2, 2p+2q
The rest except for 1000 and pqr are trinomials
It is a factorization method based on the law of distributivity which is
a(b + c) = a · b + a · c
and used in reverse as follows
a · b + a · c = a(b + c)
a is a common factor to a b and a c is therefore factored out.
Factoring a polynomial is to write it as the product of simpler polynomials.
Example:
2 x + 4 = 2(x + 2)
3 x 2 - x = x(3x - 1)
NOTE: it is very easy to check if your factorization is correct by expanding the factored form to check if you get the original polynomial
Example: check that 3 x 2 - x = x(3x - 1)
Expand x(3x - 1) using the law of distributivity.
x(3x - 1) = (x)(3x) +(x)(-1) = 3x2 - x , which is correct.
More Examples
Find a common factor and use the method of distributivity in reverse to factor the polynomials completely.
a) 9 x - 6
b) x 2 - x
c) 3 x + 12 x y
d) 16 x 3 + 8 x 2 y + 4 x y 2
e) 2 x 4(x + 5) + x 2(x + 5)
Solution to the above examples
a) Find any common factors in the two terms of 9 x - 6 by expressing both terms 9x and 6 in the given binomial as prime factorization. Hence
9 x - 6 = 3 ·3 ·x - 2 ·3
The greatest common factor is 3 and is factored out. Hence
9 x - 6 = 3 (3 x - 2)
b) The prime factorization of x 2 and x is needed to find the greatest common factor in x 2 - x.
x 2 - x = x · x - x = x · x - 1 · x
The greatest common factor is x and is therefore factored out. Hence
x 2 - x = = x (x - 1)
c) The prime factorizations of 3 x and 12 x y are needed to find the greatest common factor in 3 x + 12 x y.
3 x + 12 x y = 3 · x - 3 · 4 · x · y = 3 · x · 1 - 3 x · 4 · y
The greatest common factor is 3 x. Hence
3 x + 12 x y = 3 x (1 + 4 y)
d) The prime factorization of 16 x 3 , 8 x 2 y and 4 x y 2 are needed to find the greatest common factor in 16 x 3 + 8 x 2 y + 4 x y 2.
16 x 3 + 8 x 2 y + 4 x y 2 = 2 · 2 · 2 · 2 · x · x · x + 2 · 2 · 2 · x · x · y + 2 · 2 · x · y · y
The greatest common factor is 2 · 2 · x = 4 x. Hence
16 x 3 + 8 x 2 y + 4 x y 2 = 4 x ( 2 · 2 · x · x + 2 · x · y + y · y) = 4 x (4 x 2 + 2 x y + y 2)
e) We note that x + 5 is a common factor which can be factored out as follows:
2 x 4(x + 5) + x 2(x + 5) = (x + 5)(2 x 4 + x 2)
We now find the greatest common factor of the terms 2 x 4 and x 2 and factor it out.
2 x 4 + x 2 = 2 · x · x · x · x + x · x = x 2(2 x 2 + 1)
The complete factoring of 2 x 4(x + 5) + x 2(x + 5) is written as follows:
2 x 4(x + 5) + x 2(x + 5) = x 2(x + 5)(2 x 2 + 1)
Questions
Use common factors to factor completely the following polynomials
a) - 3 x + 9
b) 28 x + 2 x 2
c) 11 x y + 55 x 2 y
d) 20 x y + 35 x 2 y - 15 x y 2
e) 5 y (x + 1) + 10 y 2(x + 1) - 15 x y (x + 1)
Solutions to the Above Questions
a) Find any common factors in the two terms of - 3 x + 9 by expressing both terms 3 x and 9 in the given binomial as prime factorization.
- 3 x + 9 = - 3 · x - 3 · 3
The greatest common factor is 3 and is factored out. Hence
- 3x + 9 = 3 (- x + 3) = - 3 (x - 3)
b) Write the prime factorization of each of the terms in the given polynomial 28 x + 2 x 2.
28 x + 2 x 2 = 2 · 2 · 7 · x + 2 · x · x
The greatest common factor is 2 x and is factored out. Hence
28 x + 2 x 2 = 2 x (14 + x)
c) Write the prime factorization of each of the terms in the given polynomial 11 x y + 55 x 2 y.
11 x y + 55 x 2 y = 11 · x · y + 5 · 11 · x · x · y
The greatest common factor is 11 x y and is factored out. Hence
11 x y + 55 x 2 y = 11 x y(1 + 5 x)
d) Write the prime factorization of each of the terms in the given polynomial 20 x y + 35 x 2 y - 15 x y 2.
20 x y + 35 x 2 y - 15 x y 2 = 2 · 2 · 5 · x · y + 5 · 7 · x · x · y - 3 · 5 · x · y · y
The greatest common factor is 5 x y and is factored out. Hence
20 x y + 35 x 2 y - 15 x y 2 = 5 x y( 4 + 7 x - 3 y)
e) We start by factoring out the common factor (x + 1) in the given polynomial.
5 y (x + 1) + 10 y 2(x + 1) - 15 x y (x + 1) = (x + 1)(5y + 10y2 - 15 x y)
We now factor the polynomial 5y + 10y2 - 15 x y using the GCF to all three terms.
5 y + 10y2 - 15 x y = 5 · y + 2 · 5 · y · y - 3 · 5 · y · x = 5 · y (1 + 2 y - 3 x)
The given polynomial may be factored as follows.
5 y (x + 1) + 10 y 2(x + 1) - 15 x y (x + 1) = 5 y(x + 1)(1 + 2y - 3 x)