Free math worksheets

- Rewrite the divisor mentally in the form \(x-c\) to identify the synthetic value. - Use the coefficients in descending-power order. - The final synthetic entry is the remainder.

1. For divisor \(x+1=x-(-1)\), use synthetic value \(-1\) with coefficients \(1,2,-5,-6\). 2. Bring down \(1\); multiply by \(-1\) and add to get \(1\); multiply by \(-1\) and add to get \(-6\); multiply by \(-1\) and add to get remainder \(0\). 3. The quotient is \(x^2+x-6\) and the remainder is \(0\).

Quotient: \(x^2+x-6\); remainder: \(0\)

- Write a zero coefficient for every missing power before starting synthetic division. - Use the constant from \(x-1\) as the synthetic value. - Match the quotient coefficients to descending powers one degree lower than the dividend.

1. The polynomial has no \(x^3\)-term, so use coefficients \(2,0,-3,5,-4\) with synthetic value \(1\). 2. Bring down \(2\); multiply by \(1\) and add to get \(2\); multiply by \(1\) and add to get \(-1\); multiply by \(1\) and add to get \(4\); multiply by \(1\) and add to get remainder \(0\). 3. The quotient is \(2x^3+2x^2-x+4\).

Quotient: \(2x^3+2x^2-x+4\); remainder: \(0\)

- Include a zero coefficient for the missing \(x^2\)-term. - A nonzero final synthetic entry is the remainder, not another quotient coefficient. - Use the division identity to write the result after completing the synthetic work.

1. Include the missing \(x^2\)-term and use coefficients \(1,0,-4,7\) with synthetic value \(2\). 2. Bring down \(1\); multiply by \(2\) and add to get \(2\); multiply by \(2\) and add to get \(0\); multiply by \(2\) and add to get remainder \(7\). 3. The quotient is \(x^2+2x\), so \(x^3-4x+7=(x-2)(x^2+2x)+7\).

Quotient: \(x^2+2x\); remainder: \(7\) \(x^3-4x+7=(x-2)(x^2+2x)+7\)

- At each stage, divide the current leading term by the divisor's leading term. - Multiply the entire divisor by the new quotient term before subtracting. - Continue until the remaining degree is less than the divisor's degree.

1. Divide the leading terms: \(2x^3\div x=2x^2\). Multiply \(x+2\) by \(2x^2\) and subtract, leaving \(-x^2-8x-12\). 2. Divide \(-x^2\div x=-x\). Multiply \(x+2\) by \(-x\) and subtract, leaving \(-6x-12\). 3. Divide \(-6x\div x=-6\). Multiply \(x+2\) by \(-6\) and subtract, leaving remainder \(0\). 4. The quotient is \(2x^2-x-6\).

Quotient: \(2x^2-x-6\); remainder: \(0\)

- Look first at the degree and form of each divisor. - Ask whether the divisor can be represented by a single synthetic value without changing its structure. - Compare the structural requirements of the two division methods before choosing one.

a) Synthetic division is appropriate because the divisor is linear and has the form \(x-c\). Long division would also work, but it is less efficient here. b) Polynomial long division is required because the divisor has degree \(2\). Standard synthetic division in this course applies to linear divisors of the form \(x-c\).

a) Synthetic division b) Polynomial long division

- Verification reverses polynomial division. - Multiply the divisor by the proposed quotient first. - Add the proposed remainder only after expanding the product.

1. Multiply the divisor and quotient: \((x-2)(x^2+x+4)=x^3-x^2+2x-8\). 2. Add the claimed remainder: \(x^3-x^2+2x-8+5=x^3-x^2+2x-3\). 3. This matches the dividend exactly, so the quotient and remainder are correct.

The claim is correct because \((x-2)(x^2+x+4)+5=x^3-x^2+2x-3\).

- A known zero \(x=a\) gives a factor \(x-a\) and a synthetic value \(a\). - Keep the polynomial coefficients in descending-power order during synthetic division. - Factor the quadratic quotient after the division is complete. - In complete factored form, the exponent on each factor gives the multiplicity.

1. Since \(x=-1\) is a zero, divide by \(x+1\) using synthetic value \(-1\). Use coefficients \(1,-2,-13,-10\): bring down \(1\); multiply by \(-1\) and add to get \(-3\); multiply by \(-1\) and add to get \(-10\); multiply by \(-1\) and add to get remainder \(0\). The quotient is \(x^2-3x-10\). Factor the quotient: \(x^2-3x-10=(x-5)(x+2)\). Therefore, the zeros are \(x=-2\), \(x=-1\), and \(x=5\). 2. The factored form is \(f(x)=(x+1)(x-5)(x+2)\). 3. 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 the synthetic value and corresponding linear factor. - Keep the coefficients in descending-power order. - A zero remainder confirms that the known zero's factor divided evenly. - Factor the quadratic quotient to find the remaining zeros.

1. The zero \(x=-1\) corresponds to the factor \(x+1\), so use synthetic value \(-1\) with coefficients \(1,-6,3,10\). 2. Bring down \(1\). Multiply by \(-1\) and add to get \(-7\). Multiply by \(-1\) and add to get \(10\). Multiply by \(-1\) and add to get remainder \(0\). The quotient is \(x^2-7x+10\). 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 its corresponding linear factor. - Use the confirmed zero as the synthetic value. - Keep track of the leading coefficient when factoring the quadratic quotient. - Check that the final product has the same degree and leading coefficient as the original polynomial.

1. Evaluate the polynomial: \(p(1)=2-4-22+24=0\). Therefore, \(x=1\) is a zero and \(x-1\) is a factor. 2. Use synthetic value \(1\) with coefficients \(2,-4,-22,24\). Bring down \(2\); multiply by \(1\) and add to get \(-2\); multiply by \(1\) and add to get \(-24\); multiply by \(1\) and add to get remainder \(0\). The quotient is \(2x^2-2x-24\). 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 \(x=a\) gives the factor \(x-a\) and synthetic value \(a\). - Keep the coefficients in descending-power order during synthetic division. - A remainder of \(0\) confirms the known factor divides evenly. - Factor the quadratic quotient to find the remaining zeros.

a) Because \(x=3\) is a zero, \(x-3\) is a factor and the synthetic-division value is \(3\). b) Use coefficients \(1,-6,-1,30\). Bring down \(1\); multiply by \(3\) and add to get \(-3\); multiply by \(3\) and add to get \(-10\); multiply by \(3\) and add to get remainder \(0\). The quotient is \(x^2-3x-10\). Factor the quotient: \(x^2-3x-10=(x-5)(x+2)\). The remaining zeros are \(x=5\) and \(x=-2\), so \(f(x)=(x-3)(x-5)(x+2)\).

a) Use synthetic value \(3\), corresponding to factor \(x-3\). 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 \(0\). - Use one confirmed zero as the synthetic value to reduce the cubic to a quadratic. - Keep the coefficients in descending-power order. - Factor the quadratic quotient to finish finding the zeros.

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 synthetic value \(-1\) with coefficients \(1,-4,1,6\). Bring down \(1\); multiply by \(-1\) and add to get \(-5\); multiply by \(-1\) and add to get \(6\); multiply by \(-1\) and add to get remainder \(0\). The quotient is \(x^2-5x+6\). 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 each known zero as a synthetic value, one after the other. - Keep the quotient coefficients in descending-power order before the second division. - A zero remainder at each stage confirms that the corresponding linear factor divides evenly. - Factor the final quadratic quotient.

1. Use synthetic value \(-2\) with coefficients \(1,-4,-7,22,24\). Bring down \(1\); multiply by \(-2\) and add to get \(-6\); multiply by \(-2\) and add to get \(5\); multiply by \(-2\) and add to get \(12\); multiply by \(-2\) and add to get remainder \(0\). The quotient is \(x^3-6x^2+5x+12\). 2. Use synthetic value \(3\) with quotient coefficients \(1,-6,5,12\). Bring down \(1\); multiply by \(3\) and add to get \(-3\); multiply by \(3\) and add to get \(-4\); multiply by \(3\) and add to get remainder \(0\). The new quotient is \(x^2-3x-4\). 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 \(p(x) = 2(x - 1)^2(x + 3)\).

- Use each known zero as a synthetic value, one at a time. - Each zero remainder lowers the degree by \(1\) and confirms the corresponding linear factor. - Factor the remaining quadratic after the two divisions. - Check the signs in the factors: a zero \(r\) corresponds to \(x-r\).

1. Use synthetic value \(1\) with coefficients \(1,2,-7,-8,12\). Bring down \(1\); multiply by \(1\) and add to get \(3\); multiply by \(1\) and add to get \(-4\); multiply by \(1\) and add to get \(-12\); multiply by \(1\) and add to get remainder \(0\). The quotient is \(x^3+3x^2-4x-12\). 2. Use synthetic value \(-2\) with coefficients \(1,3,-4,-12\). Bring down \(1\); multiply by \(-2\) and add to get \(1\); multiply by \(-2\) and add to get \(-6\); multiply by \(-2\) and add to get remainder \(0\). The quotient is \(x^2+x-6\). 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 \(x=2\).

- Use the zero \(2\) as the synthetic value each time you test another copy of the factor \(x-2\). - Count consecutive zero remainders to determine multiplicity. - When a quotient no longer gives zero at \(x=2\), stop dividing by \(x-2\). - Factor the final quotient completely.

1. Verify the zero: \(f(2)=16-32+12+8-4=0\). 2. Use synthetic value \(2\) with coefficients \(1,-4,3,4,-4\). Bring down \(1\); multiply by \(2\) and add to get \(-2\); multiply by \(2\) and add to get \(-1\); multiply by \(2\) and add to get \(2\); multiply by \(2\) and add to get remainder \(0\). The quotient is \(x^3-2x^2-x+2\). 3. Divide again using synthetic value \(2\) with coefficients \(1,-2,-1,2\). Bring down \(1\); multiply by \(2\) and add to get \(0\); multiply by \(2\) and add to get \(-1\); multiply by \(2\) and add to get remainder \(0\). The quotient is \(x^2-1\). 4. The remaining quotient is not divisible by \(x-2\) because \(2^2-1=3\ne0\). Therefore, \(x=2\) has multiplicity \(2\). 5. Factor \(x^2-1=(x-1)(x+1)\). Thus, \(f(x)=(x-2)^2(x-1)(x+1)\).

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

- Use \(2\) as the synthetic value for each successive division. - Each zero remainder confirms one more copy of the factor \(x-2\). - After the third successful division, test the remaining factor at \(x=2\).

1. Use synthetic value \(2\) with coefficients \(1,-3,-6,28,-24\). Bring down \(1\); multiply by \(2\) and add to get \(-1\); multiply by \(2\) and add to get \(-8\); multiply by \(2\) and add to get \(12\); multiply by \(2\) and add to get remainder \(0\). The quotient is \(x^3-x^2-8x+12\). 2. Divide again with coefficients \(1,-1,-8,12\). Bring down \(1\); multiply by \(2\) and add to get \(1\); multiply by \(2\) and add to get \(-6\); multiply by \(2\) and add to get remainder \(0\). The quotient is \(x^2+x-6\). 3. Divide a third time with coefficients \(1,1,-6\). Bring down \(1\); multiply by \(2\) and add to get \(3\); multiply by \(2\) and add to get remainder \(0\). The remaining factor is \(g(x)=x+3\). 4. Since \(g(2)=5\ne0\), a fourth division by \(x-2\) would not have zero remainder. Therefore, \(x=2\) has multiplicity \(3\), not \(4\).

After three divisions, \(g(x)=x+3\). Because \(g(2)=5\ne0\), \(x=2\) has multiplicity \(3\) and not multiplicity \(4\).

- Possible integer zeros are related to factors of the constant term. - Once an integer zero is confirmed, use it as a synthetic value to lower the degree. - Repeat the process with another confirmed zero. - Solve the final quadratic equation exactly.

1. The possible integer zeros are among \(\pm1\), \(\pm2\), \(\pm5\), and \(\pm10\). Testing shows that \(f(1)=0\) and \(f(-2)=0\). 2. Use synthetic value \(1\) with coefficients \(1,1,-7,-5,10\). Bring down \(1\); multiply by \(1\) and add to get \(2\); multiply by \(1\) and add to get \(-5\); multiply by \(1\) and add to get \(-10\); multiply by \(1\) and add to get remainder \(0\). The quotient is \(x^3+2x^2-5x-10\). 3. Use synthetic value \(-2\) with coefficients \(1,2,-5,-10\). Bring down \(1\); multiply by \(-2\) and add to get \(0\); multiply by \(-2\) and add to get \(-5\); multiply by \(-2\) and add to get remainder \(0\). The quotient is \(x^2-5\). 4. Solve \(x^2-5=0\): \(x=-\sqrt{5}\) or \(x=\sqrt{5}\). 5. 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. - Use each confirmed zero as a synthetic value. - After each zero remainder, continue with the quotient coefficients. - Stop when the quotient is quadratic, then solve that equation exactly.

1. Possible integer zeros are among \(\pm1\), \(\pm2\), \(\pm3\), and \(\pm6\). Testing identifies \(x=1\), \(x=-1\), and \(x=2\) as zeros. 2. Use synthetic value \(1\) with coefficients \(1,-2,-4,8,3,-6\). Bring down \(1\); multiply and add successively to obtain \(-1,-5,3,6\), then remainder \(0\). The quotient is \(x^4-x^3-5x^2+3x+6\). 3. Use synthetic value \(-1\) with coefficients \(1,-1,-5,3,6\). Bring down \(1\); multiply and add successively to obtain \(-2,-3,6\), then remainder \(0\). The quotient is \(x^3-2x^2-3x+6\). 4. Use synthetic value \(2\) with coefficients \(1,-2,-3,6\). Bring down \(1\); multiply and add successively to obtain \(0,-3\), then remainder \(0\). The quotient is \(x^2-3\). 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 polynomial expressions equal and move all terms to one side. - After finding an integer solution, use it as the synthetic value. - Remember to include a zero coefficient for any missing power in the cubic. - Factor the quadratic quotient and then use either original function for the y-coordinates.

1. Set the functions equal: \(-x^3+3x^2=4\), so \(x^3-3x^2+4=0\). 2. Test small integer values. Since \((-1)^3-3(-1)^2+4=0\), \(x=-1\) is a solution and \(x+1\) is a factor. 3. The cubic has coefficients \(1,-3,0,4\); include the zero coefficient for the missing \(x\)-term. Use synthetic value \(-1\): bring down \(1\); multiply by \(-1\) and add to get \(-4\); multiply by \(-1\) and add to get \(4\); multiply by \(-1\) and add to get remainder \(0\). The quotient is \(x^2-4x+4\). 4. Factor the quotient: \(x^2-4x+4=(x-2)^2\). Therefore, the x-coordinates are \(x=-1\) and \(x=2\). 5. Because \(g(x)=4\), both corresponding y-coordinates are \(4\). 6. The intersection points are \((-1,4)\) and \((2,4)\).

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

- Synthetic division is not appropriate for a quadratic divisor, so use long division. - Divide leading terms to determine each new quotient term. - Subtract the full product at every stage and keep like powers aligned.

1. Divide \(2x^4\) by \(x^2\) to get \(2x^2\). Multiply the divisor by \(2x^2\) and subtract, leaving \(-x^3-4x^2-x+6\). 2. Divide \(-x^3\) by \(x^2\) to get \(-x\). Multiply the divisor by \(-x\) and subtract, leaving \(-3x^2-3x+6\). 3. Divide \(-3x^2\) by \(x^2\) to get \(-3\). Multiply the divisor by \(-3\) and subtract, leaving remainder \(0\). 4. The quotient is \(2x^2-x-3\).

Quotient: \(2x^2-x-3\); remainder: \(0\)

- Insert zero-coefficient terms so like powers stay aligned. - Long division stops when the remainder's degree is less than the divisor's degree. - Subtract the entire product at each stage, including signs.

1. Write the dividend with missing terms: \(x^4+0x^3+2x^2+0x-3\). 2. Divide \(x^4\) by \(x^2\) to get \(x^2\). Subtract \(x^2(x^2-x+1)\), leaving \(x^3+x^2-3\). 3. Divide \(x^3\) by \(x^2\) to get \(x\). Subtract \(x(x^2-x+1)\), leaving \(2x^2-x-3\). 4. Divide \(2x^2\) by \(x^2\) to get \(2\). Subtract \(2(x^2-x+1)\), leaving \(x-5\). 5. Because the remainder has degree less than \(2\), the quotient is \(x^2+x+2\) and the remainder is \(x-5\).

Quotient: \(x^2+x+2\); remainder: \(x-5\)

- Reverse the division identity to reconstruct the dividend. - Expand the divisor-quotient product before adding the remainder. - The check value comes from the zero of the divisor \(x+3\).

a) Use the division identity: \(P(x)=(x+3)(2x^2-x+4)-5\). Expanding gives \((x+3)(2x^2-x+4)=2x^3+5x^2+x+12\), so \(P(x)=2x^3+5x^2+x+7\). b) Evaluate \(P(-3)=-54+45-3+7=-5\), which matches the stated remainder.

a) \(P(x)=2x^3+5x^2+x+7\) b) \(P(-3)=-5\)

- Rewrite \(x+2\) in the form \(x-c\). - The synthetic value is the zero of the divisor, not its visible constant with the same sign. - Recheck the final remainder after correcting the setup.

1. The divisor \(x+2\) is \(x-(-2)\), so the correct synthetic value is \(-2\), not \(2\). 2. Use coefficients \(1,3,-1,-6\). Bring down \(1\); multiply by \(-2\) and add to get \(1\); multiply by \(-2\) and add to get \(-3\); multiply by \(-2\) and add to get remainder \(0\). 3. The quotient is \(x^2+x-3\), the remainder is \(0\), and \(x+2\) is a factor.

The student used the wrong synthetic value. The correct value is \(-2\), giving quotient \(x^2+x-3\) and remainder \(0\).

- Use the same synthetic value repeatedly while the remainder stays \(0\). - Each successful division removes one copy of the corresponding linear factor. - Stop when the next quotient no longer evaluates to \(0\) at the same value.

1. Use synthetic value \(1\) with coefficients \(1,-1,-3,5,-2\). The quotient coefficients are \(1,0,-3,2\) with remainder \(0\), so one factor \(x-1\) is removed. 2. Divide the quotient again by \(x-1\): using coefficients \(1,0,-3,2\) gives quotient \(x^2+x-2\) with remainder \(0\). 3. Divide once more by \(x-1\): using coefficients \(1,1,-2\) gives quotient \(x+2\) with remainder \(0\). 4. The remaining quotient has value \(3\) at \(x=1\), so it is not divisible by \(x-1\) again. Therefore, \(x=1\) has multiplicity \(3\). 5. The complete factorization is \(f(x)=(x-1)^3(x+2)\).

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

- Reverse the division identity to construct the polynomial in part a. - For part b, remember the sign convention for the divisor \(x+1\). - A remainder on division by \(x-c\) can be checked by evaluating the polynomial at \(c\).

a) Use the division identity: \(P(x)=(x-2)(x^3-3x+1)+5=x^4-2x^3-3x^2+7x+3\). b) For divisor \(x+1\), use synthetic value \(-1\) with coefficients \(1,-2,-3,7,3\). The quotient is \(x^3-3x^2+7\) and the remainder is \(-4\). c) Evaluate \(P(2)=5\), confirming the first remainder. Evaluate \(P(-1)=-4\), confirming the second remainder.

a) \(P(x)=x^4-2x^3-3x^2+7x+3\) b) Quotient \(x^3-3x^2+7\); remainder \(-4\) c) \(P(2)=5\) and \(P(-1)=-4\)