Solution - Factoring binomials using the difference of squares
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Factoring binomials using the difference of squaresStep by Step Solution
Step 1 :
Equation at the end of step 1 :
(((k12)+(19•(k6)))+88) ((k4)-(6•(k2))) ——————————————————————•————————————————————— ((5•(k8))+(40•(k2))) (((2•(k12))+19k6)-33)Step 2 :
Equation at the end of step 2 :
(((k12)+(19•(k6)))+88) ((k4)-(6•(k2))) ——————————————————————•———————————————— ((5•(k8))+(40•(k2))) ((2k12+19k6)-33)Step 3 :
Equation at the end of step 3 :
(((k12)+(19•(k6)))+88) ((k4)-(2•3k2))
——————————————————————•——————————————
((5•(k8))+(40•(k2))) (2k12+19k6-33)
Step 4 :
k4 - 6k2
Simplify ————————————————
2k12 + 19k6 - 33
Step 5 :
Pulling out like terms :
5.1 Pull out like factors :
k4 - 6k2 = k2 • (k2 - 6)
Trying to factor as a Difference of Squares :
5.2 Factoring: k2 - 6
Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)
Proof : (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 - AB + AB - B2 =
A2 - B2
Note : AB = BA is the commutative property of multiplication.
Note : - AB + AB equals zero and is therefore eliminated from the expression.
Check : 6 is not a square !!
Ruling : Binomial can not be factored as the difference of two perfect squares.
Trying to factor by splitting the middle term
5.3 Factoring 2k12 + 19k6 - 33
The first term is, 2k12 its coefficient is 2 .
The middle term is, +19k6 its coefficient is 19 .
The last term, "the constant", is -33
Step-1 : Multiply the coefficient of the first term by the constant 2 • -33 = -66
Step-2 : Find two factors of -66 whose sum equals the coefficient of the middle term, which is 19 .
| -66 | + | 1 | = | -65 | ||
| -33 | + | 2 | = | -31 | ||
| -22 | + | 3 | = | -19 | ||
| -11 | + | 6 | = | -5 | ||
| -6 | + | 11 | = | 5 | ||
| -3 | + | 22 | = | 19 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 22
2k12 - 3k6 + 22k6 - 33
Step-4 : Add up the first 2 terms, pulling out like factors :
k6 • (2k6-3)
Add up the last 2 terms, pulling out common factors :
11 • (2k6-3)
Step-5 : Add up the four terms of step 4 :
(k6+11) • (2k6-3)
Which is the desired factorization
Trying to factor as a Difference of Squares :
5.4 Factoring: 2k6-3
Check : 2 is not a square !!
Ruling : Binomial can not be factored as the
difference of two perfect squares
Polynomial Roots Calculator :
5.5 Find roots (zeroes) of : F(k) = 2k6-3
Polynomial Roots Calculator is a set of methods aimed at finding values of k for which F(k)=0
Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers k which can be expressed as the quotient of two integers
The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient
In this case, the Leading Coefficient is 2 and the Trailing Constant is -3.
The factor(s) are:
of the Leading Coefficient : 1,2
of the Trailing Constant : 1 ,3
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | -1.00 | ||||||
| -1 | 2 | -0.50 | -2.97 | ||||||
| -3 | 1 | -3.00 | 1455.00 | ||||||
| -3 | 2 | -1.50 | 19.78 | ||||||
| 1 | 1 | 1.00 | -1.00 | ||||||
| 1 | 2 | 0.50 | -2.97 | ||||||
| 3 | 1 | 3.00 | 1455.00 | ||||||
| 3 | 2 | 1.50 | 19.78 |
Polynomial Roots Calculator found no rational roots
Trying to factor as a Sum of Cubes :
5.6 Factoring: k6+11
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 : 11 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes
Polynomial Roots Calculator :
5.7 Find roots (zeroes) of : F(k) = k6+11
See theory in step 5.5
In this case, the Leading Coefficient is 1 and the Trailing Constant is 11.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,11
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 12.00 | ||||||
| -11 | 1 | -11.00 | 1771572.00 | ||||||
| 1 | 1 | 1.00 | 12.00 | ||||||
| 11 | 1 | 11.00 | 1771572.00 |
Polynomial Roots Calculator found no rational roots
Equation at the end of step 5 :
(((k12)+(19•(k6)))+88) k2•(k2-6) ——————————————————————•——————————————— ((5•(k8))+(40•(k2))) (2k6-3)•(k6+11)Step 6 :
Equation at the end of step 6 :
(((k12)+(19•(k6)))+88) k2•(k2-6) ——————————————————————•——————————————— ((5•(k8))+(23•5k2)) (2k6-3)•(k6+11)Step 7 :
Equation at the end of step 7 :
(((k12)+(19•(k6)))+88) k2•(k2-6) ——————————————————————•——————————————— (5k8+(23•5k2)) (2k6-3)•(k6+11)Step 8 :
Equation at the end of step 8 :
(((k12)+19k6)+88) k2•(k2-6)
—————————————————•———————————————
(5k8+40k2) (2k6-3)•(k6+11)
Step 9 :
k12 + 19k6 + 88
Simplify ———————————————
5k8 + 40k2
Step 10 :
Pulling out like terms :
10.1 Pull out like factors :
5k8 + 40k2 = 5k2 • (k6 + 8)
Trying to factor by splitting the middle term
10.2 Factoring k12 + 19k6 + 88
The first term is, k12 its coefficient is 1 .
The middle term is, +19k6 its coefficient is 19 .
The last term, "the constant", is +88
Step-1 : Multiply the coefficient of the first term by the constant 1 • 88 = 88
Step-2 : Find two factors of 88 whose sum equals the coefficient of the middle term, which is 19 .
| -88 | + | -1 | = | -89 | ||
| -44 | + | -2 | = | -46 | ||
| -22 | + | -4 | = | -26 | ||
| -11 | + | -8 | = | -19 | ||
| -8 | + | -11 | = | -19 | ||
| -4 | + | -22 | = | -26 | ||
| -2 | + | -44 | = | -46 | ||
| -1 | + | -88 | = | -89 | ||
| 1 | + | 88 | = | 89 | ||
| 2 | + | 44 | = | 46 | ||
| 4 | + | 22 | = | 26 | ||
| 8 | + | 11 | = | 19 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 8 and 11
k12 + 8k6 + 11k6 + 88
Step-4 : Add up the first 2 terms, pulling out like factors :
k6 • (k6+8)
Add up the last 2 terms, pulling out common factors :
11 • (k6+8)
Step-5 : Add up the four terms of step 4 :
(k6+11) • (k6+8)
Which is the desired factorization
Trying to factor as a Sum of Cubes :
10.3 Factoring: k6+11
Check : 11 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes
Polynomial Roots Calculator :
10.4 Find roots (zeroes) of : F(k) = k6+11
See theory in step 5.5
In this case, the Leading Coefficient is 1 and the Trailing Constant is 11.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,11
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 12.00 | ||||||
| -11 | 1 | -11.00 | 1771572.00 | ||||||
| 1 | 1 | 1.00 | 12.00 | ||||||
| 11 | 1 | 11.00 | 1771572.00 |
Polynomial Roots Calculator found no rational roots
Trying to factor as a Sum of Cubes :
10.5 Factoring: k6+8
Check : 8 is the cube of 2
Check : k6 is the cube of k2
Factorization is :
(k2 + 2) • (k4 - 2k2 + 4)
Trying to factor as a Sum of Cubes :
10.6 Factoring: k6 + 11
Check : 11 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes
Polynomial Roots Calculator :
10.7 Find roots (zeroes) of : F(k) = k2 + 2
See theory in step 5.5
In this case, the Leading Coefficient is 1 and the Trailing Constant is 2.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 3.00 | ||||||
| -2 | 1 | -2.00 | 6.00 | ||||||
| 1 | 1 | 1.00 | 3.00 | ||||||
| 2 | 1 | 2.00 | 6.00 |
Polynomial Roots Calculator found no rational roots
Trying to factor by splitting the middle term
10.8 Factoring k4 - 2k2 + 4
The first term is, k4 its coefficient is 1 .
The middle term is, -2k2 its coefficient is -2 .
The last term, "the constant", is +4
Step-1 : Multiply the coefficient of the first term by the constant 1 • 4 = 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 :
10.9 Factoring: k6+8
Check : 8 is the cube of 2
Check : k6 is the cube of k2
Factorization is :
(k2 + 2) • (k4 - 2k2 + 4)
Polynomial Roots Calculator :
10.10 Find roots (zeroes) of : F(k) = k2 + 2
See theory in step 5.5
In this case, the Leading Coefficient is 1 and the Trailing Constant is 2.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2
Let us test ....
| P | Q | P/Q | F(P/Q) | Divisor | |||||
|---|---|---|---|---|---|---|---|---|---|
| -1 | 1 | -1.00 | 3.00 | ||||||
| -2 | 1 | -2.00 | 6.00 | ||||||
| 1 | 1 | 1.00 | 3.00 | ||||||
| 2 | 1 | 2.00 | 6.00 |
Polynomial Roots Calculator found no rational roots
Trying to factor by splitting the middle term
10.11 Factoring k4 - 2k2 + 4
The first term is, k4 its coefficient is 1 .
The middle term is, -2k2 its coefficient is -2 .
The last term, "the constant", is +4
Step-1 : Multiply the coefficient of the first term by the constant 1 • 4 = 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
Canceling Out :
10.12 Cancel out (k2+2) which appears on both sides of the fraction line.
Canceling Out :
10.13 Cancel out (k4-2k2+4) which appears on both sides of the fraction line.
Equation at the end of step 10 :
(k6 + 11) k2 • (k2 - 6)
————————— • —————————————————————
5k2 (2k6 - 3) • (k6 + 11)
Step 11 :
Trying to factor as a Sum of Cubes :
11.1 Factoring: k6+11
Check : 11 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes
Trying to factor as a Difference of Squares :
11.2 Factoring: 2k6-3
Check : 2 is not a square !!
Ruling : Binomial can not be factored as the
difference of two perfect squares
Trying to factor as a Sum of Cubes :
11.3 Factoring: k6+11
Check : 11 is not a cube !!
Ruling : Binomial can not be factored as the difference of two perfect cubes
Canceling Out :
11.4 Cancel out (k6+11) which appears on both sides of the fraction line.
Canceling Out :
11.5 Canceling out k2 as it appears on both sides of the fraction line
Trying to factor as a Difference of Squares :
11.6 Factoring: 2k6-3
Check : 2 is not a square !!
Ruling : Binomial can not be factored as the
difference of two perfect squares
Final result :
k2 - 6
—————————————
5 • (2k6 - 3)
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