Precalculus Examples
f(x)=x2-9
Step 1
Step 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,±3,±9
q=±1
Step 1.2
Find every combination of ±pq. These are the possible roots of the polynomial function.
±1,±3,±9
±1,±3,±9
Step 2
Step 2.1
Place the numbers representing the divisor and the dividend into a division-like configuration.
3 | 1 | 0 | -9 |
Step 2.2
The first number in the dividend (1) is put into the first position of the result area (below the horizontal line).
3 | 1 | 0 | -9 |
1 |
Step 2.3
Multiply the newest entry in the result (1) by the divisor (3) and place the result of (3) under the next term in the dividend (0).
3 | 1 | 0 | -9 |
3 | |||
1 |
Step 2.4
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
3 | 1 | 0 | -9 |
3 | |||
1 | 3 |
Step 2.5
Multiply the newest entry in the result (3) by the divisor (3) and place the result of (9) under the next term in the dividend (-9).
3 | 1 | 0 | -9 |
3 | 9 | ||
1 | 3 |
Step 2.6
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
3 | 1 | 0 | -9 |
3 | 9 | ||
1 | 3 | 0 |
Step 2.7
All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.
(1)x+3
Step 2.8
Simplify the quotient polynomial.
x+3
x+3
Step 3
Since 3>0 and all of the signs in the bottom row of the synthetic division are positive, 3 is an upper bound for the real roots of the function.
Upper Bound: 3
Step 4
Step 4.1
Place the numbers representing the divisor and the dividend into a division-like configuration.
-3 | 1 | 0 | -9 |
Step 4.2
The first number in the dividend (1) is put into the first position of the result area (below the horizontal line).
-3 | 1 | 0 | -9 |
1 |
Step 4.3
Multiply the newest entry in the result (1) by the divisor (-3) and place the result of (-3) under the next term in the dividend (0).
-3 | 1 | 0 | -9 |
-3 | |||
1 |
Step 4.4
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
-3 | 1 | 0 | -9 |
-3 | |||
1 | -3 |
Step 4.5
Multiply the newest entry in the result (-3) by the divisor (-3) and place the result of (9) under the next term in the dividend (-9).
-3 | 1 | 0 | -9 |
-3 | 9 | ||
1 | -3 |
Step 4.6
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
-3 | 1 | 0 | -9 |
-3 | 9 | ||
1 | -3 | 0 |
Step 4.7
All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.
(1)x-3
Step 4.8
Simplify the quotient polynomial.
x-3
x-3
Step 5
Since -3<0 and the signs in the bottom row of the synthetic division alternate sign, -3 is a lower bound for the real roots of the function.
Lower Bound: -3
Step 6
Step 6.1
Place the numbers representing the divisor and the dividend into a division-like configuration.
9 | 1 | 0 | -9 |
Step 6.2
The first number in the dividend (1) is put into the first position of the result area (below the horizontal line).
9 | 1 | 0 | -9 |
1 |
Step 6.3
Multiply the newest entry in the result (1) by the divisor (9) and place the result of (9) under the next term in the dividend (0).
9 | 1 | 0 | -9 |
9 | |||
1 |
Step 6.4
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
9 | 1 | 0 | -9 |
9 | |||
1 | 9 |
Step 6.5
Multiply the newest entry in the result (9) by the divisor (9) and place the result of (81) under the next term in the dividend (-9).
9 | 1 | 0 | -9 |
9 | 81 | ||
1 | 9 |
Step 6.6
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
9 | 1 | 0 | -9 |
9 | 81 | ||
1 | 9 | 72 |
Step 6.7
All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.
(1)x+9+72x-9
Step 6.8
Simplify the quotient polynomial.
x+9+72x-9
x+9+72x-9
Step 7
Since 9>0 and all of the signs in the bottom row of the synthetic division are positive, 9 is an upper bound for the real roots of the function.
Upper Bound: 9
Step 8
Step 8.1
Place the numbers representing the divisor and the dividend into a division-like configuration.
-9 | 1 | 0 | -9 |
Step 8.2
The first number in the dividend (1) is put into the first position of the result area (below the horizontal line).
-9 | 1 | 0 | -9 |
1 |
Step 8.3
Multiply the newest entry in the result (1) by the divisor (-9) and place the result of (-9) under the next term in the dividend (0).
-9 | 1 | 0 | -9 |
-9 | |||
1 |
Step 8.4
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
-9 | 1 | 0 | -9 |
-9 | |||
1 | -9 |
Step 8.5
Multiply the newest entry in the result (-9) by the divisor (-9) and place the result of (81) under the next term in the dividend (-9).
-9 | 1 | 0 | -9 |
-9 | 81 | ||
1 | -9 |
Step 8.6
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
-9 | 1 | 0 | -9 |
-9 | 81 | ||
1 | -9 | 72 |
Step 8.7
All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.
(1)x-9+72x+9
Step 8.8
Simplify the quotient polynomial.
x-9+72x+9
x-9+72x+9
Step 9
Since -9<0 and the signs in the bottom row of the synthetic division alternate sign, -9 is a lower bound for the real roots of the function.
Lower Bound: -9
Step 10
Determine the upper and lower bounds.
Upper Bounds: 3,9
Lower Bounds: -3,-9
Step 11