Solution - Factoring binomials using the difference of squares
Step by Step Solution
Step 1 :
Equation at the end of step 1 :
((8 • (x6)) + (33•37x3)) - 125Step 2 :
Equation at the end of step 2 :
(23x6 + (33•37x3)) - 125
Step 3 :
Trying to factor by splitting the middle term
3.1 Factoring 8x6+999x3-125
The first term is, 8x6 its coefficient is 8 .
The middle term is, +999x3 its coefficient is 999 .
The last term, "the constant", is -125
Step-1 : Multiply the coefficient of the first term by the constant 8 • -125 = -1000
Step-2 : Find two factors of -1000 whose sum equals the coefficient of the middle term, which is 999 .
| -1000 | + | 1 | = | -999 | ||
| -500 | + | 2 | = | -498 | ||
| -250 | + | 4 | = | -246 | ||
| -200 | + | 5 | = | -195 | ||
| -125 | + | 8 | = | -117 | ||
| -100 | + | 10 | = | -90 | ||
| -50 | + | 20 | = | -30 | ||
| -40 | + | 25 | = | -15 | ||
| -25 | + | 40 | = | 15 | ||
| -20 | + | 50 | = | 30 | ||
| -10 | + | 100 | = | 90 | ||
| -8 | + | 125 | = | 117 | ||
| -5 | + | 200 | = | 195 | ||
| -4 | + | 250 | = | 246 | ||
| -2 | + | 500 | = | 498 | ||
| -1 | + | 1000 | = | 999 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -1 and 1000
8x6 - 1x3 + 1000x3 - 125
Step-4 : Add up the first 2 terms, pulling out like factors :
x3 • (8x3-1)
Add up the last 2 terms, pulling out common factors :
125 • (8x3-1)
Step-5 : Add up the four terms of step 4 :
(x3+125) • (8x3-1)
Which is the desired factorization
Trying to factor as a Difference of Cubes:
3.2 Factoring: 8x3-1
Theory : A difference of two perfect cubes, a3 - b3 can be factored into
(a-b) • (a2 +ab +b2)
Proof : (a-b)•(a2+ab+b2) =
a3+a2b+ab2-ba2-b2a-b3 =
a3+(a2b-ba2)+(ab2-b2a)-b3 =
a3+0+0-b3 =
a3-b3
Check : 8 is the cube of 2
Check : 1 is the cube of 1
Check : x3 is the cube of x1
Factorization is :
(2x - 1) • (4x2 + 2x + 1)
Trying to factor by splitting the middle term
3.3 Factoring 4x2 + 2x + 1
The first term is, 4x2 its coefficient is 4 .
The middle term is, +2x its coefficient is 2 .
The last term, "the constant", is +1
Step-1 : Multiply the coefficient of the first term by the constant 4 • 1 = 4
Step-2 : Find two factors of 4 whose sum equals the coefficient of the middle term, which is 2 .
| -4 | + | -1 | = | -5 | ||
| -2 | + | -2 | = | -4 | ||
| -1 | + | -4 | = | -5 | ||
| 1 | + | 4 | = | 5 | ||
| 2 | + | 2 | = | 4 | ||
| 4 | + | 1 | = | 5 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Trying to factor as a Sum of Cubes :
3.4 Factoring: x3+125
Theory : A sum of two perfect cubes, a3 + b3 can be factored into :
(a+b) • (a2-ab+b2)
Proof : (a+b) • (a2-ab+b2) =
a3-a2b+ab2+ba2-b2a+b3 =
a3+(a2b-ba2)+(ab2-b2a)+b3=
a3+0+0+b3=
a3+b3
Check : 125 is the cube of 5
Check : x3 is the cube of x1
Factorization is :
(x + 5) • (x2 - 5x + 25)
Trying to factor by splitting the middle term
3.5 Factoring x2 - 5x + 25
The first term is, x2 its coefficient is 1 .
The middle term is, -5x its coefficient is -5 .
The last term, "the constant", is +25
Step-1 : Multiply the coefficient of the first term by the constant 1 • 25 = 25
Step-2 : Find two factors of 25 whose sum equals the coefficient of the middle term, which is -5 .
| -25 | + | -1 | = | -26 | ||
| -5 | + | -5 | = | -10 | ||
| -1 | + | -25 | = | -26 | ||
| 1 | + | 25 | = | 26 | ||
| 5 | + | 5 | = | 10 | ||
| 25 | + | 1 | = | 26 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Final result :
(2x-1)•(4x2+2x+1)•(x+5)•(x2-5x+25)
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