Andrew M. answered • 05/15/15
Tutor
New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors
1) For the first one we have 9x2 - 4
Note that this is actually the difference of two squares: (3x)2 - 22
The formula for difference of two squares is a2 - b2 = (a-b)(a+b)
In this case a = 3x, b = 2
So 9x2 - 4 = (3x-2)(3x+2)
2) For 216x3 - 1 we have the difference of two cubes since (6x)3 = 216x3 and 13 = 1
The formula for difference of two cubes is a3 - b3 = (a-b)(a2 + ab + b2)
In this case a = 6x, b = 1
Thus 216x3 - 1 = (6x-1)[(6x)2 + 6x(1) + 12]
= (6x-1)(36x2 + 6x + 1)
3) 200x2 - 50
We can factor 50 out of both terms giving
200x2 - 50 = 50(4x2-1)
Note that 4x2 - 1 is the difference of two squares... (2x)2 - (1)2 so again use a2-b2 = (a-b)(a+b)
So 4x2 - 1 = (2x-1)(2x+1)
Thus we have:
200x2-50 = 50(2x-1)(2x+1)
4) 128x3 - 2
Factor out a 2 giving
2(64x3 - 1)
64x3 - 1 is the difference of two cubes ... 64x3-1 = (4x)3 - (1)3
Again use the forumula for difference of two cubes a3-b3 = (a-b)(a2+ab+b2) where a = 4x, b = 1
64x3 - 1 = (4x-1)[(4x)2 + (4x)(1) + (1)2]
= (4x-1)(16x2 + 4x + 1)
Don't forget that this is still multiplied by the 2 that we factored out so:
128x3 -2 = 2(4x-1)(16x2 + 4x + 1)
5) 3x3 + 81
Let's factor out the 3 giving
3x3 + 81 = 3(x3 + 27)
Note that x3 + 27 is the sum of two cubes x3 + 27 = (x)3 + (3)3
The formula for factoring sum of two cubes is a3 + b3 = (a+b)(a2 - ab + b2) .... a = x, b = 3
Thus (x3 + 27) = (x+3)(x2 -3x +32) = (x+3)(x2-3x+9)
Don't forget the factor of 3 we took out at the beginning.... so,
3x3 + 81 = 3(x+3)(x2 - 3x +9)
6) x3 + x2 + x + 1
I note that I can create two sets of "x+1" if I factor an x2 out of the first two terms giving:
x3 + x2 + x + 1 = x2(x+1) + x + 1
If I insert a set of parentheses to make the grouping obvious I have:
x2(x+1) + 1(x+1)
Note that ab + cb = (a+c)(b) so I have
x2(x+1) + 1(x+1) = (x2 + 1)(x+1)
x3 + x2 + x + 1 = (x2 +1)(x + 1)
7) x3 + 2x2 + 5x + 10
I look for factors and see that I can create 2 sets of "x+2" if I factor x2 out of the x3+2x2 and 5 out of x+10
x3 + 2x2 + 5x + 10 = x2(x+2) + 5(x+2)
We have the same situation as in the previous factor where ab + cb = (a+b)(c) so
x2 + 2x + 5x + 10 = (x2+5)(x+2)
8) 9x3 + 18x2 + 7x + 14
Again we can factor to create two sets of "x+2"
9x3 + 18x2 + 7x + 14 = 9x2(x+2) + 7(x+2) = (9x2 + 7)(x + 2)
9) 2x3 + 4x2 + 4x + 8
The trick is the same as the last couple... factor 2x2 from the first 2 terms and factor 4 from the next 2 terms
to create two sets of (x+2) ... 2x2(x + 2) + 4(x+2) = (2x2 + 4)(x+2)= 2(x2 + 2)(x + 2)
10) 2x3 -2x2 + 5x - 5
Create two sets of (x-1) by factoring the two first terms and the two 2nd terms and do as above
11) 5x3 - 20x2 + 3x -12
Create two sets of (x-4) by factoring the 1st two terms and the 2nd two terms
12) 7x3 + 14x2 + 7x
We can factor 7x out of every term giving:
7x(x2 + 2x + 1)
Note that x2 + 2x + 1 fits the pattern of (a + b)2 = a2 + 2ab + b2 where a = x, b = 1
x2 + 2x + 1 = (x+1)2 so ...
7x3 + 14x2 + 7x = 7x(x + 1)2
13) 2x3 - 4x2 - 3x + 6
as in problems 6 thru 11, I can create two sets of (x-2) by factoring
2x(x-2) -3(x-2) = (2x-3)(x-2)
14) 3x3 - 24 = 3(x3 - 8)
Again we have the difference of two cubes since x3 - 8 = (x)3 - (2)3
a3 - b3 = (a-b)(a2 +ab + b2) ... a = x, b = 2
3x3 - 24 = 3(x-2)(x2 + 2x + 4)
15) 3x4 - 300x2
We can factor 3x2 from each term giving
3x4 - 300x2 = 3x2(x2 - 100)
Note that x2 - 100 is the difference of two squares... (x)2 - (10)2
and a2 - b2 = (a-b)(a+b) so...
3x4 - 300x2 = 3x2(x-10)(x+10)
16) 3x4 + 3x3 + 6x2 + 6x
I can again factor the 1st two terms and 2nd two terms to create sets of (x+1)
3x4 + 3x3 + 6x2 + 6x = 3x3(x + 1) + 6x(x + 1)
= (3x3 + 6x)(x+1)
Note we can still factor 3x out of the first parentheses giving
3x(x2 + 2)(x+1)
17) 10x3 - 20x2 - 2x + 4
As in problem 16 I can factor the 1st two terms and 2nd two terms to create sets of (x-2)
10x3 - 20x2 - 2x + 4 = 10x2(x-2) - 2(x-2) = (10x2 - 2)(x-2)
Again... note we can factor a 2 out of the the first parentheses giving
2(5x2 - 1)(x-2)
