Free Math Worksheets

- A known zero \(x = a\) gives a factor \(x - a\). - Divide the polynomial by the known linear factor to reduce its degree. - Factor the resulting quadratic expression. - In complete factored form, the exponent on each factor gives the multiplicity.

1. Because \(x = -1\) is a zero, \(x + 1\) is a factor. Dividing gives \((x^3 - 2x^2 - 13x - 10) \div (x + 1) = x^2 - 3x - 10\). 2. Factor the quotient: \(x^2 - 3x - 10 = (x - 5)(x + 2)\). Therefore, the zeros are \(x = -2\), \(x = -1\), and \(x = 5\). 3. The factored form is \(f(x) = (x + 1)(x - 5)(x + 2)\). 4. Each linear factor occurs once, so each zero has multiplicity 1.

1. \(x = -2\), \(x = -1\), and \(x = 5\) 2. \(f(x) = (x + 1)(x - 5)(x + 2)\) 3. Each zero has multiplicity 1.

- A known zero identifies a corresponding linear factor. - Divide the cubic polynomial by that linear factor to reduce the degree. - Polynomial division should produce a quadratic quotient and remainder \(0\). - Factor or otherwise solve the resulting quadratic equation.

1. The zero \(x = -1\) corresponds to the factor \(x + 1\). 2. Divide: \((x^3 - 6x^2 + 3x + 10) \div (x + 1) = x^2 - 7x + 10\), with remainder \(0\). 3. Factor the quotient: \(x^2 - 7x + 10 = (x - 2)(x - 5)\). 4. Therefore, the remaining zeros are \(x = 2\) and \(x = 5\).

\(x = 2\) and \(x = 5\)

- Substituting a value that produces \(0\) confirms a zero and a linear factor. - Keep track of the leading coefficient during division and factoring. - A cubic with three real zeros factors into a constant times three linear factors. - Factor the quadratic quotient completely.

1. Evaluate the polynomial: \(p(1) = 2 - 4 - 22 + 24 = 0\). Therefore, \(x = 1\) is a zero, and \(x - 1\) is a factor. 2. Divide: \((2x^3 - 4x^2 - 22x + 24) \div (x - 1) = 2x^2 - 2x - 24\), with remainder \(0\). 3. Factor the quotient: \(2x^2 - 2x - 24 = 2(x^2 - x - 12) = 2(x - 4)(x + 3)\). 4. Therefore, the complete factorization is \(p(x) = 2(x - 1)(x - 4)(x + 3)\).

\(p(x) = 2(x - 1)(x - 4)(x + 3)\)

- A known zero determines a corresponding linear factor. - Use polynomial division or synthetic division to reduce the degree. - The quotient will be a quadratic expression. - Factor or solve that quadratic to find the remaining zeros.

1. Because \(x = 3\) is a zero, the factor theorem shows that \(x - 3\) is a factor. Divide \(f(x)\) by \(x - 3\) using polynomial division or synthetic division. 2. The division gives \((x^3 - 6x^2 - x + 30) \div (x - 3) = x^2 - 3x - 10\), with remainder \(0\). 3. Factor the quotient: \(x^2 - 3x - 10 = (x - 5)(x + 2)\). The remaining zeros are \(x = 5\) and \(x = -2\). 4. Therefore, \(f(x) = (x - 3)(x - 5)(x + 2)\).

a) Divide \(f(x)\) by the factor \(x - 3\), then factor the resulting quadratic quotient. b) Remaining zeros: \(x = -2\) and \(x = 5\) Factored form: \(f(x) = (x - 3)(x - 5)(x + 2)\)

- Divide by the linear factor associated with the known zero. - Check whether the quadratic quotient is a perfect-square trinomial. - A repeated solution of the quadratic creates a repeated zero of the original polynomial. - Preserve the original leading coefficient in the final factorization.

1. Divide by the factor corresponding to the known zero: \((4x^3 - 8x^2 - 11x - 3) \div (x - 3) = 4x^2 + 4x + 1\), with remainder \(0\). 2. Factor the quotient: \(4x^2 + 4x + 1 = (2x + 1)^2\). Therefore, the additional zero is \(x = -\frac{1}{2}\). 3. The complete factorization is \(g(x) = (x - 3)(2x + 1)^2\), equivalently \(g(x) = 4(x - 3)\left(x + \frac{1}{2}\right)^2\). 4. The zero \(x = 3\) has multiplicity 1, and \(x = -\frac{1}{2}\) has multiplicity 2.

Zeros: \(x = 3\) with multiplicity 1; \(x = -\frac{1}{2}\) with multiplicity 2 Factored form: \(g(x) = (x - 3)(2x + 1)^2\)

- Substitute each proposed value and check whether the output is \(0\). - Once you know a zero, divide by its corresponding linear factor. - Polynomial or synthetic division reduces the cubic to a quadratic. - Factor or solve the quadratic quotient.

1. Evaluate the proposed values: \(f(1) = 4\), \(f(-1) = 0\), \(f(2) = 0\), and \(f(-2) = -20\). Therefore, \(x = -1\) and \(x = 2\) are zeros among the tested values. 2. Use one known zero to reduce the degree. Dividing by \(x + 1\) gives \((x^3 - 4x^2 + x + 6) \div (x + 1) = x^2 - 5x + 6\), with remainder \(0\). 3. Factor the quotient: \(x^2 - 5x + 6 = (x - 2)(x - 3)\). 4. Thus, all zeros are \(x = -1\), \(x = 2\), and \(x = 3\).

Tested zeros: \(x = -1\) and \(x = 2\) All zeros: \(x = -1\), \(x = 2\), and \(x = 3\)

- Test the small integers in the stated interval. - Divide by the linear factor corresponding to the zero you find. - Check whether the quadratic quotient is a perfect-square trinomial. - Read multiplicities from the exponents in the complete factorization.

1. Testing integers gives \(g(1) = 1 - 5 + 8 - 4 = 0\), so \(x = 1\) is a zero. 2. Divide by the corresponding factor: \((x^3 - 5x^2 + 8x - 4) \div (x - 1) = x^2 - 4x + 4\), with remainder \(0\). 3. Factor the quotient: \(x^2 - 4x + 4 = (x - 2)^2\). 4. Therefore, \(g(x) = (x - 1)(x - 2)^2\). The zero \(x = 1\) has multiplicity 1, and \(x = 2\) has multiplicity 2.

Factored form: \(g(x) = (x - 1)(x - 2)^2\) Zeros: \(x = 1\) with multiplicity 1; \(x = 2\) with multiplicity 2

- Use polynomial division once for each known zero to reduce the quartic to a quadratic. - Pay attention to the signs in the corresponding linear factors. - The order in which you use the two known zeros does not affect the final quotient. - Factor or solve the resulting quadratic.

1. Divide by the factor corresponding to \(x = -2\): \((x^4 - 4x^3 - 7x^2 + 22x + 24) \div (x + 2) = x^3 - 6x^2 + 5x + 12\), with remainder \(0\). 2. Divide the quotient by the factor corresponding to \(x = 3\): \((x^3 - 6x^2 + 5x + 12) \div (x - 3) = x^2 - 3x - 4\), with remainder \(0\). 3. Factor the quadratic: \(x^2 - 3x - 4 = (x - 4)(x + 1)\). 4. Therefore, the remaining zeros are \(x = 4\) and \(x = -1\).

\(x = -1\) and \(x = 4\)

- Try small integer values to find one zero of the cubic polynomial. - Possible integer zeros are related to the factors of the constant term. - After finding a zero \(x_0\), divide by \(x - x_0\) to reduce the degree. - Factor or solve the quadratic quotient to find the remaining zeros.

1. Test possible integer zeros among the factors of the constant term \(10\). Since \(h(1) = 1 - 4 - 7 + 10 = 0\), \(x = 1\) is a zero. 2. Divide the polynomial by its corresponding factor: \((x^3 - 4x^2 - 7x + 10) \div (x - 1) = x^2 - 3x - 10\), with remainder \(0\). 3. Factor the quotient: \(x^2 - 3x - 10 = (x - 5)(x + 2)\). 4. Therefore, \(h(x) = (x - 1)(x - 5)(x + 2)\), so the zeros are \(x = -2\), \(x = 1\), and \(x = 5\).

The zeros are \(x = -2\), \(x = 1\), and \(x = 5\).

- Substitute each given number into the function and check whether the result is zero. - After finding a zero \(x_0\), divide the polynomial by \(x - x_0\). - Polynomial division reduces the cubic to a quadratic. - Factor or solve the quadratic quotient to find the remaining zeros.

1. Evaluate the function at each given number: \(f(-3) = (-3)^3 - 7 \cdot (-3) - 6 = -12 \ne 0\), \(f(-1) = (-1)^3 - 7 \cdot (-1) - 6 = 0\), and \(f(0.5) = (0.5)^3 - 7 \cdot 0.5 - 6 = -9.375 \ne 0\). Therefore, \(x = -1\) is the zero among the given numbers. 2. Divide by the corresponding factor: \((x^3 - 7x - 6) \div (x + 1) = x^2 - x - 6\), with remainder \(0\). 3. Factor the quotient: \(x^2 - x - 6 = (x - 3)(x + 2)\). 4. Thus, \(f(x) = (x + 1)(x - 3)(x + 2)\), and the remaining zeros are \(x = -2\) and \(x = 3\).

Among the given numbers, \(-1\) is a zero. The remaining zeros are \(x = -2\) and \(x = 3\).

- Substitute each given number and check whether the function value is zero. - Use polynomial division or synthetic division after identifying one zero. - Be careful with signs when dividing by a factor such as \(x + 1\). - Factor or solve the quadratic quotient, including its leading coefficient.

1. Evaluate the function at each given number: \(g(-1) = 2 \cdot (-1)^3 - 5 \cdot (-1)^2 - 4 \cdot (-1) + 3 = 0\), \(g(2) = 2 \cdot 2^3 - 5 \cdot 2^2 - 4 \cdot 2 + 3 = -9 \ne 0\), and \(g(1.5) = 2 \cdot (1.5)^3 - 5 \cdot (1.5)^2 - 4 \cdot 1.5 + 3 = -7.5 \ne 0\). Therefore, \(x = -1\) is the zero among the given numbers. 2. Divide by the corresponding factor: \((2x^3 - 5x^2 - 4x + 3) \div (x + 1) = 2x^2 - 7x + 3\), with remainder \(0\). 3. Factor the quotient: \(2x^2 - 7x + 3 = (2x - 1)(x - 3)\). 4. Thus, the remaining zeros are \(x = 0.5\) and \(x = 3\).

Among the given numbers, \(-1\) is a zero. The remaining zeros are \(x = 0.5\) and \(x = 3\).

- Possible integer zeros are related to the factors of the constant term. - After finding one zero, divide by its corresponding linear factor. - Polynomial or synthetic division reduces the cubic to a quadratic. - A zero \(x = r\) corresponds to the factor \(x - r\).

1. Test possible integer zeros among the factors of \(-12\). Since \(f(-1) = -1 + 2 + 11 - 12 = 0\), \(x = -1\) is a zero. 2. Divide by the corresponding factor: \((x^3 + 2x^2 - 11x - 12) \div (x + 1) = x^2 + x - 12\), with remainder \(0\). 3. Factor the quotient: \(x^2 + x - 12 = (x - 3)(x + 4)\). 4. Therefore, \(f(x) = (x + 1)(x - 3)(x + 4)\), and the zeros are \(x = -4\), \(x = -1\), and \(x = 3\).

The zeros are \(x = -4\), \(x = -1\), and \(x = 3\). The linear-factor form is \(f(x) = (x + 1)(x - 3)(x + 4)\).

- Test small integer factors of the constant term to find one zero. - Divide by the linear factor associated with the known zero. - A repeated linear factor indicates a zero with multiplicity greater than 1.

1. Test possible integer zeros among the factors of \(18\). Since \(g(3) = 27 - 36 - 9 + 18 = 0\), \(x = 3\) is a zero. 2. Divide by the corresponding factor: \((x^3 - 4x^2 - 3x + 18) \div (x - 3) = x^2 - x - 6\), with remainder \(0\). 3. Factor the quotient: \(x^2 - x - 6 = (x - 3)(x + 2)\). 4. Therefore, \(g(x) = (x - 3)^2(x + 2)\). The zero \(x = 3\) has multiplicity 2, and the zero \(x = -2\) has multiplicity 1.

\(x = 3\) with multiplicity 2; \(x = -2\) with multiplicity 1

- Clear the fractions to make testing possible integer zeros easier. - After finding a zero \(x_0\), divide by \(x - x_0\). - Keep the leading coefficient when writing the complete factorization. - Expand the factors to check the result.

1. Multiply the equation \(f(x) = 0\) by \(4\) to work with \(x^3 - 6x^2 + 11x - 6 = 0\). Testing integer factors of \(-6\) shows that \(x = 1\) is a zero. 2. Divide the original polynomial by \(x - 1\): \(f(x) \div (x - 1) = \frac{1}{4}x^2 - \frac{5}{4}x + \frac{3}{2}\), with remainder \(0\). 3. Factor the quadratic quotient: \(\frac{1}{4}x^2 - \frac{5}{4}x + \frac{3}{2} = \frac{1}{4}(x^2 - 5x + 6) = \frac{1}{4}(x - 2)(x - 3)\). 4. Therefore, \(f(x) = \frac{1}{4}(x - 1)(x - 2)(x - 3)\), so the zeros are \(x = 1\), \(x = 2\), and \(x = 3\).

The zeros are \(x = 1\), \(x = 2\), and \(x = 3\). The linear-factor form is \(f(x) = \frac{1}{4}(x - 1)(x - 2)(x - 3)\).

- Substitute \(x = 1\) to verify the given zero. - Divide by \(x - 1\) to reduce the cubic to a quadratic. - Check whether the known zero appears again when the quadratic is factored. - Multiplicity is shown by the exponent on a linear factor.

1. Evaluate the function: \(p(1) = 2 + 2 - 10 + 6 = 0\). Therefore, \(x = 1\) is a zero. 2. Divide by the corresponding factor: \((2x^3 + 2x^2 - 10x + 6) \div (x - 1) = 2x^2 + 4x - 6\), with remainder \(0\). 3. Factor the quotient: \(2x^2 + 4x - 6 = 2(x^2 + 2x - 3) = 2(x - 1)(x + 3)\). 4. Therefore, \(p(x) = 2(x - 1)^2(x + 3)\). The zero \(x = 1\) has multiplicity 2, and the zero \(x = -3\) has multiplicity 1.

The zeros are \(x = 1\) with multiplicity 2 and \(x = -3\) with multiplicity 1. The linear-factor form is \(p(x) = 2(x - 1)^2(x + 3)\).

- Divide successively by the linear factors associated with the known zeros. - Each successful division lowers the degree by \(1\). - Factor the remaining quadratic after the two divisions. - Check the signs in the factors: a zero \(r\) corresponds to \(x - r\).

1. Divide by the factor associated with \(x = 1\): \((x^4 + 2x^3 - 7x^2 - 8x + 12) \div (x - 1) = x^3 + 3x^2 - 4x - 12\), with remainder \(0\). 2. Divide the quotient by the factor associated with \(x = -2\): \((x^3 + 3x^2 - 4x - 12) \div (x + 2) = x^2 + x - 6\), with remainder \(0\). 3. Factor the remaining quadratic: \(x^2 + x - 6 = (x - 2)(x + 3)\). 4. Therefore, \(p(x) = (x - 1)(x + 2)(x - 2)(x + 3)\). The zeros are \(x = -3\), \(x = -2\), \(x = 1\), and \(x = 2\).

The complete factorization is \(p(x) = (x - 1)(x + 2)(x - 2)(x + 3)\). The zeros are \(x = -3\), \(x = -2\), \(x = 1\), and \(x = 2\).

- Count how many times division by \(x - 2\) produces a zero remainder. - Evaluate the final quotient at \(x = 2\) to decide whether another division is possible. - Factor the remaining quadratic expression completely.

1. Verify the zero: \(f(2) = 16 - 32 + 12 + 8 - 4 = 0\). 2. Divide once: \((x^4 - 4x^3 + 3x^2 + 4x - 4) \div (x - 2) = x^3 - 2x^2 - x + 2\), with remainder \(0\). 3. Divide again: \((x^3 - 2x^2 - x + 2) \div (x - 2) = x^2 - 1\), with remainder \(0\). 4. The remaining quotient is not divisible by \(x - 2\) because \(2^2 - 1 = 3 \ne 0\). Therefore, \(x = 2\) has multiplicity 2. 5. Factor the difference of squares: \(x^2 - 1 = (x - 1)(x + 1)\). Thus, \(f(x) = (x - 2)^2(x - 1)(x + 1)\).

The zero \(x = 2\) has multiplicity 2. The complete factorization is \(f(x) = (x - 2)^2(x - 1)(x + 1)\).

- Perform the polynomial divisions one at a time. - Each of the first three divisions must have remainder \(0\). - Evaluate the remaining factor at \(x = 2\) to test whether a fourth division is possible.

1. Divide once: \((x^4 - 3x^3 - 6x^2 + 28x - 24) \div (x - 2) = x^3 - x^2 - 8x + 12\), with remainder \(0\). 2. Divide again: \((x^3 - x^2 - 8x + 12) \div (x - 2) = x^2 + x - 6\), with remainder \(0\). 3. Divide a third time: \((x^2 + x - 6) \div (x - 2) = x + 3\), with remainder \(0\). 4. The remaining factor is \(g(x) = x + 3\). 5. Since \(g(2) = 5 \ne 0\), the remaining factor is not divisible by \(x - 2\). Therefore, \(x = 2\) has multiplicity 3, not 4.

After three divisions, the remaining factor is \(g(x) = x + 3\). Because \(g(2) = 5 \ne 0\), \(x = 2\) has multiplicity 3 and not multiplicity 4.

- Possible integer zeros are related to the factors of the constant term. - Use each known zero to divide out its corresponding linear factor. - A degree-\(4\) polynomial can have at most \(4\) real zeros. - Continue dividing until the remaining equation is quadratic.

1. The possible integer zeros are among \(\pm 1\), \(\pm 2\), \(\pm 5\), and \(\pm 10\). 2. Testing shows that \(f(1) = 0\) and \(f(-2) = 0\), so \(x = 1\) and \(x = -2\) are zeros. 3. Divide by \(x - 1\): \((x^4 + x^3 - 7x^2 - 5x + 10) \div (x - 1) = x^3 + 2x^2 - 5x - 10\), with remainder \(0\). 4. Divide the quotient by \(x + 2\): \((x^3 + 2x^2 - 5x - 10) \div (x + 2) = x^2 - 5\), with remainder \(0\). 5. Solve \(x^2 - 5 = 0\): \(x = -\sqrt{5}\) or \(x = \sqrt{5}\). 6. Therefore, the zeros are \(x = -\sqrt{5}\), \(x = -2\), \(x = 1\), and \(x = \sqrt{5}\).

The zeros are \(x = -\sqrt{5}\), \(x = -2\), \(x = 1\), and \(x = \sqrt{5}\).

- Start by testing integer factors of the constant term. - A zero \(a\) allows division by \(x - a\). - Repeat the process until the quotient is quadratic. - Each successful division should have remainder \(0\).

1. Possible integer zeros are among \(\pm 1\), \(\pm 2\), \(\pm 3\), and \(\pm 6\). Testing identifies \(x = 1\), \(x = -1\), and \(x = 2\) as zeros. 2. Divide by \(x - 1\): \(f(x) \div (x - 1) = x^4 - x^3 - 5x^2 + 3x + 6\), with remainder \(0\). 3. Divide by \(x + 1\): \((x^4 - x^3 - 5x^2 + 3x + 6) \div (x + 1) = x^3 - 2x^2 - 3x + 6\), with remainder \(0\). 4. Divide by \(x - 2\): \((x^3 - 2x^2 - 3x + 6) \div (x - 2) = x^2 - 3\), with remainder \(0\). 5. Solve \(x^2 - 3 = 0\): \(x = -\sqrt{3}\) or \(x = \sqrt{3}\). 6. Therefore, the zeros are \(x = -\sqrt{3}\), \(x = -1\), \(x = 1\), \(x = \sqrt{3}\), and \(x = 2\).

The zeros are \(x = -\sqrt{3}\), \(x = -1\), \(x = 1\), \(x = \sqrt{3}\), and \(x = 2\).

- Set the two function expressions equal to find intersection points. - Try small integer values to find one solution of the cubic equation. - After finding one linear factor, use polynomial division to reduce the degree. - Use either function to find the y-coordinate for each solution.

1. Set the functions equal: \(-x^3 + 3x^2 = 4\). 2. Rewrite the equation as \(x^3 - 3x^2 + 4 = 0\). 3. Test small integer values. Since \((-1)^3 - 3(-1)^2 + 4 = 0\), \(x = -1\) is a solution and \(x + 1\) is a factor. 4. Divide: \((x^3 - 3x^2 + 4) \div (x + 1) = x^2 - 4x + 4\). 5. Factor the quotient: \(x^2 - 4x + 4 = (x - 2)^2\). Therefore, the x-coordinates are \(x = -1\) and \(x = 2\). 6. Because \(g(x) = 4\), both corresponding y-coordinates are \(4\). 7. The intersection points are \((-1, 4)\) and \((2, 4)\).

The intersection points are \((-1, 4)\) and \((2, 4)\).