| Synthetic Division/Rational Roots |
-- Section 5.3 -- |
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Use synthetic division to determine if (x - 1) is a factor of P(x) = x3 - 2x2 - 5x + 6 |
| Put 1 in the upper left hand corner (box). Put the coefficients of the polynomial in the first row. | ||
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Solve |
Step |
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Bring the down the 1. |
Use 1 (in box) if x -
1 is the factor. For x - r , use r |
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Multiply
1 × 1
= 1
Add Columns -2 + 1 = -1 |
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Multiply
-1 × 1
= 1
Add Columns -5 + (-1) = -6 |
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Multiply
-6 × 1
= -6
Add Columns 6 + (-6) = 0 |
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If there is a remainder of zero, 0 , then 1 is a zero and (x -1) is a factor of the polynomial. |
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Use synthetic division to determine if (x - 1) is a factor of P(x) = 2x3 - 2x2 - 5x + 6 |
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[Solution] |
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Find all rational roots of P(x) = 2x3 + 3x2 - 14x - 15 |
| Find all possible rational zeros - rational numbers are ratio of integers i.e. -2/1 = -2, 0, 1/2, -3/2 | ||
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Solve |
Step |
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p = ± 1, ±
3, ± 5, ±
15
q ± 1, ± 2 |
Find
possible rational zeros p/q where p are factors of the last coefficient 15. q are the factors of the first coefficient 2. |
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p = ± 1, ± 2, ± 3, ± 6 and |
p/1 = p and all of the p/2's. | We always get the p's |
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Try the easy ones first. | |
| Start with -1 (I tried 1 first but it failed - You can use Descartes' Rules of Signs to narrow the search.) | ||
| Put -1 in the upper left hand corner (box). Put the coefficients of the polynomial in the first row. | ||
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Bring the down the 2. | coefficient of 2x3 is 2 |
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Multiply
2 × -1
= -2
Add Columns 3 + (-2) = 1 |
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Multiply
1 × -1
= -1 Add Columns -14 + (-1) = -15 |
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Multiply
-15 × -1
= 15
Add Columns -15 + 15 = 0 |
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If there is a remainder of zero, 0 , then -1 is a zero and (x -(-1)) = (x + 1) is a factor of the polynomial. |
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Q(x) = 2x2 + x - 15 |
The quotient is the coefficients of the bottom line | Once we have a quadratic equation, factor or use the quadratic formula. |
2x2 + x - 15 = (2x - 1)(x + 3) = 0 |
Set equal to zero and solve for x | |
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2x - 1 = 0 or x + 3 = 0 |
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x = 1/2 or x = -3 |
Check in original equation. | |
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{ -1, 1/2, -3} |
There are three solutions for P(x) | P(1/2) = P(-1) = P(-3) = 0 |
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Side note: P(x) = 2x3 + 3x2 - 14x - 15 = (x + 1)(2x - 1)(x + 3) |
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Find all possible rational roots and all rational roots of P(x) = 4x3 - 15x2 - 31x + 30 |
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[Solution] |
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Find all roots of P(x) = 3x4 + 10x3 - 9x2 - 40x - 12 |
| Find all possible rational zeros - rational numbers are ratio of integers i.e. -2/1 = -2, 0, 1/2, -3/2 | ||
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Solve |
Step |
Property |
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p = ± 1, ±
2, ± 3, ±
4, ± 6, ±
12
q ± 1, ± 3 |
Find
possible rational zeros p/q where p are factors of the last coefficient 12. q are the factors of the first coefficient 2. |
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p = ± 1, ± 2, ± 3, ± 4, ± 6, ± 12 and |
p/1 = p | We always get the p's |
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and all of the p/3's. | |
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Reduce | |
| Start with 2 (I tried others first - You can use Descartes' Rules of Signs to narrow the search.) | ||
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Bring the down the 3. | coefficient of 3x4 is 3 |
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Multiply
3 × 2
= 6 Add Columns 10 + 6 = 16 |
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Multiply
16 × 2
= 32 Add Columns -9 + 32 = 23 |
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Multiply
23 × 2
= 46 Add Columns -40 + 46 = 6 |
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Multiply
6 × 2
= 12 Add Columns -12 + 12 = 0 |
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| If there is a remainder of zero, 0 , then 2 is a zero and (x - 2) is a factor of the polynomial. | ||
Q(x) = 3x3 + 16x2 + 23x + 6 |
The quotient is the coefficients of the bottom line | |
| Try -2 next | I tried other numbers that failed first | |
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I did the work for you.... | |
| If there is a remainder of zero, 0 , then 2 is a zero and (x - (-2)) = (x + 2) is a factor of the polynomial. | ||
| Q(x) = 3x2 + 10x + 3 | The quotient is the coefficients of the bottom line | Once we have a quadratic equation, factor or use the quadratic formula. |
3x2 + 10x + 35 = (3x + 1)(x + 3) = 0 |
Set equal to zero and solve for x | |
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3x + 1 = 0 or x + 3 = 0 |
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x = -1/3 or x = -3 |
Check in original equation. | |
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{ 2, -2, -1/3 , -3} |
There are three solutions for P(x) | P(2) = P(-2) = P(-1/3) = P(-3) = 0 |
| Side note: P(x) = 3x4 + 10x3 - 9x2 - 40x - 12 = (x - 2)(x + 2)(3x + 1)(x + 3) | ||
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Tutorials and Applets by
Joe McDonald
Community College of Southern Nevada
