factoring polynomials

Question

Did I factor this polynomial correctly?

 

1.  25x^2 - 36y^2 =

   =  (5x)^2 - (6y)^2

   = (5x + 6y)(5x - 6y)

 

 

would you help  me with these factoring polynomials.

 

1.  x^2 + x - 6

 

2.  6x^2 + 13x - 5

 

3.  4x^4 - 4x^2 + 1

 

thanks

 

Answer 1

When attempting to factor a trinomial, sometimes you can’t seem to find the right numbers and you’re left wondering if you’re just overlooking something. Here’s a method that will always find the right factors over the integers, if they exist, or prove that the trinomial is not factorable over the integers. We use this pattern:

(ax+b)(cx+d) = (ac)x^2 + (ad+bc)x + (bd)

The product of the leading coefficient, (ac), and the constant term, (bd), is (abcd), which can be factored into (ad)(bc). The coefficient of x, (ad+bc), is the sum of those two factors.

6x^2 + 13x - 5

Find the product of the leading coefficient and the constant term:
–5(6) = –30

Factors | Sum =? 13
-1(30)  | 29
–2(15) | 13 <== use these

6x^2 – 2x + 15x - 5

2x(3x – 1) + 5(3x - 1)
(3x – 1)(2x + 5)

This method will also prove that a trinomial is NOT factorable over the integers; e.g.,:

2x^2 - 3x - 18

2(-18) = -36

Factors | Sum =? –3
1(-36)  | -35
2(-18)  | -16
3(-12)  | -9
4(-9)    | -5
6(-6)    | 0
-1(36)  | 35
-2(18)  | 16
-3(12)  | 9
-4(9)    | 5
-6(6)    | 0

No possible pair of integer factors works, so trinomial is not factorable over the integers.

Let’s find factors:

Apply <h,k> to (y=ax^2) ––> y–k = a(x–h)^2
a(x-h)^2 + k <== Vertex Form
ax^2 - 2ahx + ah^2 + k <== Expanded
2x^2 - 3x - 18 <== Standard Form

Standard Form : Expanded Vertex Form:
2 = a
3=2ah=4h, h = 3/4
-18=ah^2+k=2(3/4)^2+k=9/8+k, k=-144/8-9/8=-153/8

2(x-3/4)^2 - 153/8 = 0
(x-3/4)^2 = 153/8/2 = 9*17/16
|x-3/4| = 3√(17)/4; b/c √(a^2) == |a|
x = 3/4 ± 3√(17)/4

So Factored Form of 2x^2 - 3x - 18 is:
2 (x – 3/4 – 3√(17)/4) (x – 3/4 + 3√(17)/4).

Check:
2(x^2 - 3/4 x + x 3√(17)/4
          - 3/4 x + 9/16          – 9√(17)/16
                     – x 3√(17)/4 + 9√(17)/16 - 9*17/16)
2(x^2 - 3/2 x - 16*9/16)
2x^2 - 3 x - 18 √

====

4x^4 - 4x^2 + 1 = (2x^2)^2 - 2(2x^2) + 1,
which is a Perfect Square Trinomial
= (2x^2 – 1)^2 = ((x√(2))^2 – 1^2)^2,
now use Difference of Squares,
= (x√(2) – 1)^2(x√(2) + 1)^2.

Answer 2

#1 is correct. If you ever want to find out if you have factored something properly, just FOIL it out again. In this case, the middle terms are -30xy+30xy, which cancel, leaving (25x²-36y²).

 

For the others;

 

x²+x-6, We have to find out what two numbers, when multiplied together equal -6, but when added or subtracted equals 1. 6*1 equals 6, but when added or subtracted equals 5 or -5. So these don't work.

 

6x²+13x-5, I'm not sure if there is some secrete to doing this other than plug and play. Since there is only 5*1 that will equal 5, we must also figure out which two numbers, when multiplied equal 6. In this case, 3 and 2 are used. Now you just have to play with it to figure out which order they go in. Try each of these;

 

(3x+5)(2x-1), (2x+5)(3x-1), (3x+1)(2x-5). There are many more, but your answer in these three!

 

For 4x⁴-4x²+1, use a simple substitution to figure it out. Let x⁴=a², then a=x² this just becomes 4a²-4a+1, which is easier to factor. 

Remember, the original problem is in terms of x, so make sure you plug back in the appropriate values. In other words, you answer may look something like;

 

(Aa+B)(Ca+D), plug in x² for a to get (Ax²+B)(Cx²+D).

Answer 3

Your first answer is correct.

I think factoring out these types of problems is mostly guess and check

 

ax²+bx+c

Get factor pairs from a and factor pairs from c, and try to pair them together to get b

 

1. x^2 + x - 6

a: 1,1

c:1,6 or 2,3   because c is negative, one number from the pair must be negative

we have to get 1 for b

1*3 + 1*(-2) = 1

 

(x+3) (x-2) 

 

2. a: 1,6 or 2,3

c: 1,5 (again one number must be negative)

get 13 for b

 

3.

This is one step up from the usual format, but we can do it the same way

a: 1,4 or 2,2

c: 1,1 (c is positive so either both are negative or both are positive)

b: -4

guess:

1*(-1) + 4*(-1) = -5   this doesn't work

2*(-1) + 2*(-1) = -4  this matches b, this works

 

(2x²-1) (2x²-1)

(2x²-1)²

We can now deal with these the same way you did your first problem

((root(2)*x)²-1²)    ²

((root2*x + 1)(root2*x-1))²

(root2*x+1)²(root2*x-1)²

 

Answer 4

1. You can check the answer you provided through the FOIL (multiple the first two nomials, the outer two nomials, the inner two nomials, the last two nomials) method:

(5x+6y)(5x-6y) = (5x*5x) + (5x*-6y) + (6y*5x) + (6y*-6y) = 25x^2 + -30xy + 30xy + -36y^2 = 25x^2 - 36y^2

 

So, you're correct!

 

Following polynomials:

 

1. x^2 + x - 6

 

The factors of -6 are (2,-3), (-2,3), (6,-1), and (-6,1). Trying each of these options reveals that (-2,3) works best:

 

x^2 + x - 6 = (x-2)(x+3)

 

Check your work through the FOIL method: (x*x) + (3*x) + (-2*x) + (-2*3) = x^2 + x - 6

 

2. 6x^2 + 13x - 5

 

The factors of -5 are (-5,1) and (-1,5). To factor correctly, you must factor the 6x^2 as well as the -5. The factors of 6 are (2,3) and (6,1). To find the answer, you must combine the factors of -5 and 6 to find an answer that will result in the 13x.

 

Trying options reveals that the best answer is (3x-1)(2x+5). 

 

Check your work through the FOIL method: (3x*2x) + (3x*5) + (2x*-1) + (-1*5) = 6x^2 + 15x + (-2x) + (-5) = 6x^2 + 13x - 5

 

3. 4x^4 - 4x^2 + 1

 

The factors of 1 are (-1,-1) and (1,1). You must also factor the 4 in 4x^4. The factors for 4 are (2,2), (4,1), (-2,-2), and (-4,-1).To find the answer, you must combine the factors of 1 and the factors of 4 to find an answer that will result in the -4x^2.

 

Trying options reveals that the best answer is (2x^2 - 1)(2x^2 - 1), also written as (2x^2 - 1)^2.

 

Check your work with FOIL: (2x^2*2x^2) + (2x^2*-1) + (2x^2*-1) + (-1*-1) = (4x^4) + (-2x^2) + (-2x^2) + 1 = 4x^4 - 4x^2 + 1.

Answer 5

1) yes

 

1) x^2+x-6=(x+3)(x-2)

 

2) 6x^2+13x-5=

    6*5=30 and 15*2=30 and 15-2=13

    6x^2+13x-5=(3x-1)(2x+5)

 

3) 4x^4-4x^2+1=(2x^2-1)(2x^2-1)