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Algebra Examples
x4-5x3+5x2+5x-6x4−5x3+5x2+5x−6
Step 1
Regroup terms.
-5x3+5x2+x4+5x-6
Step 2
Step 2.1
Factor -5x2 out of -5x3.
-5x2(x)+5x2+x4+5x-6
Step 2.2
Factor -5x2 out of 5x2.
-5x2(x)-5x2(-1)+x4+5x-6
Step 2.3
Factor -5x2 out of -5x2(x)-5x2(-1).
-5x2(x-1)+x4+5x-6
-5x2(x-1)+x4+5x-6
Step 3
Step 3.1
If a polynomial function has integer coefficients, then every rational zero will have the form pq where p is a factor of the constant and q is a factor of the leading coefficient.
p=±1,±6,±2,±3
q=±1
Step 3.2
Find every combination of ±pq. These are the possible roots of the polynomial function.
±1,±6,±2,±3
Step 3.3
Substitute 1 and simplify the expression. In this case, the expression is equal to 0 so 1 is a root of the polynomial.
Step 3.3.1
Substitute 1 into the polynomial.
14+5⋅1-6
Step 3.3.2
Raise 1 to the power of 4.
1+5⋅1-6
Step 3.3.3
Multiply 5 by 1.
1+5-6
Step 3.3.4
Add 1 and 5.
6-6
Step 3.3.5
Subtract 6 from 6.
0
0
Step 3.4
Since 1 is a known root, divide the polynomial by x-1 to find the quotient polynomial. This polynomial can then be used to find the remaining roots.
x4+5x-6x-1
Step 3.5
Divide x4+5x-6 by x-1.
Step 3.5.1
Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of 0.
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 |
Step 3.5.2
Divide the highest order term in the dividend x4 by the highest order term in divisor x.
x3 | |||||||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 |
Step 3.5.3
Multiply the new quotient term by the divisor.
x3 | |||||||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
+ | x4 | - | x3 |
Step 3.5.4
The expression needs to be subtracted from the dividend, so change all the signs in x4-x3
x3 | |||||||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 |
Step 3.5.5
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x3 | |||||||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 |
Step 3.5.6
Pull the next terms from the original dividend down into the current dividend.
x3 | |||||||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 |
Step 3.5.7
Divide the highest order term in the dividend x3 by the highest order term in divisor x.
x3 | + | x2 | |||||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 |
Step 3.5.8
Multiply the new quotient term by the divisor.
x3 | + | x2 | |||||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
+ | x3 | - | x2 |
Step 3.5.9
The expression needs to be subtracted from the dividend, so change all the signs in x3-x2
x3 | + | x2 | |||||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 |
Step 3.5.10
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x3 | + | x2 | |||||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 |
Step 3.5.11
Pull the next terms from the original dividend down into the current dividend.
x3 | + | x2 | |||||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 | + | 5x |
Step 3.5.12
Divide the highest order term in the dividend x2 by the highest order term in divisor x.
x3 | + | x2 | + | x | |||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 | + | 5x |
Step 3.5.13
Multiply the new quotient term by the divisor.
x3 | + | x2 | + | x | |||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 | + | 5x | ||||||||||
+ | x2 | - | x |
Step 3.5.14
The expression needs to be subtracted from the dividend, so change all the signs in x2-x
x3 | + | x2 | + | x | |||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 | + | 5x | ||||||||||
- | x2 | + | x |
Step 3.5.15
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x3 | + | x2 | + | x | |||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 | + | 5x | ||||||||||
- | x2 | + | x | ||||||||||
+ | 6x |
Step 3.5.16
Pull the next terms from the original dividend down into the current dividend.
x3 | + | x2 | + | x | |||||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 | + | 5x | ||||||||||
- | x2 | + | x | ||||||||||
+ | 6x | - | 6 |
Step 3.5.17
Divide the highest order term in the dividend 6x by the highest order term in divisor x.
x3 | + | x2 | + | x | + | 6 | |||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 | + | 5x | ||||||||||
- | x2 | + | x | ||||||||||
+ | 6x | - | 6 |
Step 3.5.18
Multiply the new quotient term by the divisor.
x3 | + | x2 | + | x | + | 6 | |||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 | + | 5x | ||||||||||
- | x2 | + | x | ||||||||||
+ | 6x | - | 6 | ||||||||||
+ | 6x | - | 6 |
Step 3.5.19
The expression needs to be subtracted from the dividend, so change all the signs in 6x-6
x3 | + | x2 | + | x | + | 6 | |||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 | + | 5x | ||||||||||
- | x2 | + | x | ||||||||||
+ | 6x | - | 6 | ||||||||||
- | 6x | + | 6 |
Step 3.5.20
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x3 | + | x2 | + | x | + | 6 | |||||||
x | - | 1 | x4 | + | 0x3 | + | 0x2 | + | 5x | - | 6 | ||
- | x4 | + | x3 | ||||||||||
+ | x3 | + | 0x2 | ||||||||||
- | x3 | + | x2 | ||||||||||
+ | x2 | + | 5x | ||||||||||
- | x2 | + | x | ||||||||||
+ | 6x | - | 6 | ||||||||||
- | 6x | + | 6 | ||||||||||
0 |
Step 3.5.21
Since the remander is 0, the final answer is the quotient.
x3+x2+x+6
x3+x2+x+6
Step 3.6
Write x4+5x-6 as a set of factors.
-5x2(x-1)+(x-1)(x3+x2+x+6)
-5x2(x-1)+(x-1)(x3+x2+x+6)
Step 4
Step 4.1
Factor x3+x2+x+6 using the rational roots test.
Step 4.1.1
If a polynomial function has integer coefficients, then every rational zero will have the form pq where p is a factor of the constant and q is a factor of the leading coefficient.
p=±1,±6,±2,±3
q=±1
Step 4.1.2
Find every combination of ±pq. These are the possible roots of the polynomial function.
±1,±6,±2,±3
Step 4.1.3
Substitute -2 and simplify the expression. In this case, the expression is equal to 0 so -2 is a root of the polynomial.
Step 4.1.3.1
Substitute -2 into the polynomial.
(-2)3+(-2)2-2+6
Step 4.1.3.2
Raise -2 to the power of 3.
-8+(-2)2-2+6
Step 4.1.3.3
Raise -2 to the power of 2.
-8+4-2+6
Step 4.1.3.4
Add -8 and 4.
-4-2+6
Step 4.1.3.5
Subtract 2 from -4.
-6+6
Step 4.1.3.6
Add -6 and 6.
0
0
Step 4.1.4
Since -2 is a known root, divide the polynomial by x+2 to find the quotient polynomial. This polynomial can then be used to find the remaining roots.
x3+x2+x+6x+2
Step 4.1.5
Divide x3+x2+x+6 by x+2.
Step 4.1.5.1
Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of 0.
x | + | 2 | x3 | + | x2 | + | x | + | 6 |
Step 4.1.5.2
Divide the highest order term in the dividend x3 by the highest order term in divisor x.
x2 | |||||||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 |
Step 4.1.5.3
Multiply the new quotient term by the divisor.
x2 | |||||||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
+ | x3 | + | 2x2 |
Step 4.1.5.4
The expression needs to be subtracted from the dividend, so change all the signs in x3+2x2
x2 | |||||||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 |
Step 4.1.5.5
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2 | |||||||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 |
Step 4.1.5.6
Pull the next terms from the original dividend down into the current dividend.
x2 | |||||||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 | + | x |
Step 4.1.5.7
Divide the highest order term in the dividend -x2 by the highest order term in divisor x.
x2 | - | x | |||||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 | + | x |
Step 4.1.5.8
Multiply the new quotient term by the divisor.
x2 | - | x | |||||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 | + | x | ||||||||
- | x2 | - | 2x |
Step 4.1.5.9
The expression needs to be subtracted from the dividend, so change all the signs in -x2-2x
x2 | - | x | |||||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 | + | x | ||||||||
+ | x2 | + | 2x |
Step 4.1.5.10
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2 | - | x | |||||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 | + | x | ||||||||
+ | x2 | + | 2x | ||||||||
+ | 3x |
Step 4.1.5.11
Pull the next terms from the original dividend down into the current dividend.
x2 | - | x | |||||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 | + | x | ||||||||
+ | x2 | + | 2x | ||||||||
+ | 3x | + | 6 |
Step 4.1.5.12
Divide the highest order term in the dividend 3x by the highest order term in divisor x.
x2 | - | x | + | 3 | |||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 | + | x | ||||||||
+ | x2 | + | 2x | ||||||||
+ | 3x | + | 6 |
Step 4.1.5.13
Multiply the new quotient term by the divisor.
x2 | - | x | + | 3 | |||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 | + | x | ||||||||
+ | x2 | + | 2x | ||||||||
+ | 3x | + | 6 | ||||||||
+ | 3x | + | 6 |
Step 4.1.5.14
The expression needs to be subtracted from the dividend, so change all the signs in 3x+6
x2 | - | x | + | 3 | |||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 | + | x | ||||||||
+ | x2 | + | 2x | ||||||||
+ | 3x | + | 6 | ||||||||
- | 3x | - | 6 |
Step 4.1.5.15
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2 | - | x | + | 3 | |||||||
x | + | 2 | x3 | + | x2 | + | x | + | 6 | ||
- | x3 | - | 2x2 | ||||||||
- | x2 | + | x | ||||||||
+ | x2 | + | 2x | ||||||||
+ | 3x | + | 6 | ||||||||
- | 3x | - | 6 | ||||||||
0 |
Step 4.1.5.16
Since the remander is 0, the final answer is the quotient.
x2-x+3
x2-x+3
Step 4.1.6
Write x3+x2+x+6 as a set of factors.
-5x2(x-1)+(x-1)((x+2)(x2-x+3))
-5x2(x-1)+(x-1)((x+2)(x2-x+3))
Step 4.2
Remove unnecessary parentheses.
-5x2(x-1)+(x-1)(x+2)(x2-x+3)
-5x2(x-1)+(x-1)(x+2)(x2-x+3)
Step 5
Step 5.1
Factor x-1 out of -5x2(x-1).
(x-1)(-5x2)+(x-1)(x+2)(x2-x+3)
Step 5.2
Factor x-1 out of (x-1)(x+2)(x2-x+3).
(x-1)(-5x2)+(x-1)((x+2)(x2-x+3))
Step 5.3
Factor x-1 out of (x-1)(-5x2)+(x-1)((x+2)(x2-x+3)).
(x-1)(-5x2+(x+2)(x2-x+3))
(x-1)(-5x2+(x+2)(x2-x+3))
Step 6
Expand (x+2)(x2-x+3) by multiplying each term in the first expression by each term in the second expression.
(x-1)(-5x2+x⋅x2+x(-x)+x⋅3+2x2+2(-x)+2⋅3)
Step 7
Step 7.1
Multiply x by x2 by adding the exponents.
Step 7.1.1
Multiply x by x2.
Step 7.1.1.1
Raise x to the power of 1.
(x-1)(-5x2+x1x2+x(-x)+x⋅3+2x2+2(-x)+2⋅3)
Step 7.1.1.2
Use the power rule aman=am+n to combine exponents.
(x-1)(-5x2+x1+2+x(-x)+x⋅3+2x2+2(-x)+2⋅3)
(x-1)(-5x2+x1+2+x(-x)+x⋅3+2x2+2(-x)+2⋅3)
Step 7.1.2
Add 1 and 2.
(x-1)(-5x2+x3+x(-x)+x⋅3+2x2+2(-x)+2⋅3)
(x-1)(-5x2+x3+x(-x)+x⋅3+2x2+2(-x)+2⋅3)
Step 7.2
Rewrite using the commutative property of multiplication.
(x-1)(-5x2+x3-x⋅x+x⋅3+2x2+2(-x)+2⋅3)
Step 7.3
Multiply x by x by adding the exponents.
Step 7.3.1
Move x.
(x-1)(-5x2+x3-(x⋅x)+x⋅3+2x2+2(-x)+2⋅3)
Step 7.3.2
Multiply x by x.
(x-1)(-5x2+x3-x2+x⋅3+2x2+2(-x)+2⋅3)
(x-1)(-5x2+x3-x2+x⋅3+2x2+2(-x)+2⋅3)
Step 7.4
Move 3 to the left of x.
(x-1)(-5x2+x3-x2+3x+2x2+2(-x)+2⋅3)
Step 7.5
Multiply -1 by 2.
(x-1)(-5x2+x3-x2+3x+2x2-2x+2⋅3)
Step 7.6
Multiply 2 by 3.
(x-1)(-5x2+x3-x2+3x+2x2-2x+6)
(x-1)(-5x2+x3-x2+3x+2x2-2x+6)
Step 8
Add -x2 and 2x2.
(x-1)(-5x2+x3+x2+3x-2x+6)
Step 9
Subtract 2x from 3x.
(x-1)(-5x2+x3+x2+x+6)
Step 10
Add -5x2 and x2.
(x-1)(x3-4x2+x+6)
Step 11
Step 11.1
Rewrite x3-4x2+x+6 in a factored form.
Step 11.1.1
Factor x3-4x2+x+6 using the rational roots test.
Step 11.1.1.1
If a polynomial function has integer coefficients, then every rational zero will have the form pq where p is a factor of the constant and q is a factor of the leading coefficient.
p=±1,±6,±2,±3
q=±1
Step 11.1.1.2
Find every combination of ±pq. These are the possible roots of the polynomial function.
±1,±6,±2,±3
Step 11.1.1.3
Substitute -1 and simplify the expression. In this case, the expression is equal to 0 so -1 is a root of the polynomial.
Step 11.1.1.3.1
Substitute -1 into the polynomial.
(-1)3-4(-1)2-1+6
Step 11.1.1.3.2
Raise -1 to the power of 3.
-1-4(-1)2-1+6
Step 11.1.1.3.3
Raise -1 to the power of 2.
-1-4⋅1-1+6
Step 11.1.1.3.4
Multiply -4 by 1.
-1-4-1+6
Step 11.1.1.3.5
Subtract 4 from -1.
-5-1+6
Step 11.1.1.3.6
Subtract 1 from -5.
-6+6
Step 11.1.1.3.7
Add -6 and 6.
0
0
Step 11.1.1.4
Since -1 is a known root, divide the polynomial by x+1 to find the quotient polynomial. This polynomial can then be used to find the remaining roots.
x3-4x2+x+6x+1
Step 11.1.1.5
Divide x3-4x2+x+6 by x+1.
Step 11.1.1.5.1
Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of 0.
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 |
Step 11.1.1.5.2
Divide the highest order term in the dividend x3 by the highest order term in divisor x.
x2 | |||||||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 |
Step 11.1.1.5.3
Multiply the new quotient term by the divisor.
x2 | |||||||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
+ | x3 | + | x2 |
Step 11.1.1.5.4
The expression needs to be subtracted from the dividend, so change all the signs in x3+x2
x2 | |||||||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 |
Step 11.1.1.5.5
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2 | |||||||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 |
Step 11.1.1.5.6
Pull the next terms from the original dividend down into the current dividend.
x2 | |||||||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 | + | x |
Step 11.1.1.5.7
Divide the highest order term in the dividend -5x2 by the highest order term in divisor x.
x2 | - | 5x | |||||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 | + | x |
Step 11.1.1.5.8
Multiply the new quotient term by the divisor.
x2 | - | 5x | |||||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 | + | x | ||||||||
- | 5x2 | - | 5x |
Step 11.1.1.5.9
The expression needs to be subtracted from the dividend, so change all the signs in -5x2-5x
x2 | - | 5x | |||||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 | + | x | ||||||||
+ | 5x2 | + | 5x |
Step 11.1.1.5.10
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2 | - | 5x | |||||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 | + | x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 6x |
Step 11.1.1.5.11
Pull the next terms from the original dividend down into the current dividend.
x2 | - | 5x | |||||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 | + | x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 6x | + | 6 |
Step 11.1.1.5.12
Divide the highest order term in the dividend 6x by the highest order term in divisor x.
x2 | - | 5x | + | 6 | |||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 | + | x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 6x | + | 6 |
Step 11.1.1.5.13
Multiply the new quotient term by the divisor.
x2 | - | 5x | + | 6 | |||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 | + | x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 6x | + | 6 | ||||||||
+ | 6x | + | 6 |
Step 11.1.1.5.14
The expression needs to be subtracted from the dividend, so change all the signs in 6x+6
x2 | - | 5x | + | 6 | |||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 | + | x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 6x | + | 6 | ||||||||
- | 6x | - | 6 |
Step 11.1.1.5.15
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2 | - | 5x | + | 6 | |||||||
x | + | 1 | x3 | - | 4x2 | + | x | + | 6 | ||
- | x3 | - | x2 | ||||||||
- | 5x2 | + | x | ||||||||
+ | 5x2 | + | 5x | ||||||||
+ | 6x | + | 6 | ||||||||
- | 6x | - | 6 | ||||||||
0 |
Step 11.1.1.5.16
Since the remander is 0, the final answer is the quotient.
x2-5x+6
x2-5x+6
Step 11.1.1.6
Write x3-4x2+x+6 as a set of factors.
(x-1)((x+1)(x2-5x+6))
(x-1)((x+1)(x2-5x+6))
Step 11.1.2
Factor x2-5x+6 using the AC method.
Step 11.1.2.1
Factor x2-5x+6 using the AC method.
Step 11.1.2.1.1
Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is 6 and whose sum is -5.
-3,-2
Step 11.1.2.1.2
Write the factored form using these integers.
(x-1)((x+1)((x-3)(x-2)))
(x-1)((x+1)((x-3)(x-2)))
Step 11.1.2.2
Remove unnecessary parentheses.
(x-1)((x+1)(x-3)(x-2))
(x-1)((x+1)(x-3)(x-2))
(x-1)((x+1)(x-3)(x-2))
Step 11.2
Remove unnecessary parentheses.
(x-1)(x+1)(x-3)(x-2)
(x-1)(x+1)(x-3)(x-2)