Free Math Worksheets

- Substitute the proposed zero into the expression. - Group the first two terms and the last two terms to reveal a common binomial factor. - Consider whether any real number has a square equal to \(-1\).

1. Substitute \(x = 2\): \(T(2) = 2^3 - 2(2^2) + 2 - 2 = 8 - 8 + 2 - 2 = 0\). Therefore, \(x = 2\) is a zero. 2. Factor by grouping: \(x^3 - 2x^2 + x - 2 = x^2(x - 2) + 1(x - 2)\). 3. Factor out \(x - 2\): \(T(x) = (x - 2)(x^2 + 1)\). 4. The equation \(x^2 + 1 = 0\) would require \(x^2 = -1\), which has no real solution. Therefore, \(x^2 + 1\) has no real linear factors.

a) \(T(2) = 0\) b) \(T(x) = (x - 2)(x^2 + 1)\) c) \(x^2 + 1 > 0\) for every real \(x\), so it has no real zeros and no real linear factors.

- Represent the zeros as \(r\) and \(3r\). - Use the zeros to write the monic quadratic as a product of two linear factors. - Expand the factors and compare coefficients with the given quadratic.

1. Let the zeros be \(r\) and \(3r\). 2. Because the quadratic is monic, its factored form is \((x-r)(x-3r)\). 3. Expand: \((x-r)(x-3r)=x^2-4rx+3r^2\). 4. Compare the constant term with \(x^2+px+12\): \(3r^2=12\), so \(r^2=4\) and \(r=2\) or \(r=-2\). 5. Compare the coefficient of \(x\): \(p=-4r\). Thus \(p=-8\) when \(r=2\), and \(p=8\) when \(r=-2\).

\(p = -8\) or \(p = 8\)

- Use the given zero in the polynomial to determine \(q\). - After finding \(q\), factor the resulting quadratic. - Set each linear factor equal to zero to identify the zeros.

1. Since \(x = -1\) is a zero, \(P(-1) = 0\). 2. Substitute \(x = -1\): \((-1)^2 + 5(-1) + q = 0\). 3. Simplify: \(1 - 5 + q = 0\), so \(q = 4\). 4. Then \(P(x) = x^2 + 5x + 4 = (x + 1)(x + 4)\). 5. Therefore, the zeros are \(x = -1\) and \(x = -4\).

\(q = 4\); \(P(x) = (x + 1)(x + 4)\); zeros: \(x = -1\) and \(x = -4\)

- Use the factor theorem to connect the factor \(x + 2\) with the zero \(-2\). - Substitute the zero to determine \(b\). - Divide or compare coefficients to find the remaining factor.

1. If \(x + 2\) is a factor, then \(x = -2\) is a zero. 2. Substitute \(x = -2\): \(3(-2)^2 + b(-2) - 10 = 0\). 3. Simplify: \(12 - 2b - 10 = 0\), so \(b = 1\). 4. The expression is \(3x^2 + x - 10\). Since one factor is \(x + 2\), the other factor is \(3x - 5\). 5. Therefore, \(3x^2 + x - 10 = (x + 2)(3x - 5)\).

1. \(b = 1\) 2. \((x + 2)(3x - 5)\)

- For the first expression, add and subtract a term that creates a perfect square. - Test small integer values to look for a zero of the cubic. - Once a zero is found, use polynomial or synthetic division. - Check whether the resulting quadratic is a perfect-square trinomial.

1. Rewrite the first expression by completing a square: \(x^4 + 3x^2 + 4 = (x^4 + 4x^2 + 4) - x^2 = (x^2 + 2)^2 - x^2\). 2. Apply the difference-of-squares pattern: \((x^2 + 2 - x)(x^2 + 2 + x) = (x^2 - x + 2)(x^2 + x + 2)\). Each quadratic has discriminant \(-7\), so neither factors further over the real numbers. 3. For the second expression, test \(z = -2\): \((-2)^3 - 6(-2)^2 + 32 = 0\). Thus, \(z + 2\) is a factor. 4. Polynomial division gives \((z^3 - 6z^2 + 32) \div (z + 2) = z^2 - 8z + 16 = (z - 4)^2\). 5. Therefore, the complete factorization is \((z + 2)(z - 4)^2\).

1) \((x^2 - x + 2)(x^2 + x + 2)\) 2) \((z + 2)(z - 4)^2\)