3.2 Quadratic & polynomial equations

Quadratic and Polynomial Equations

A quadratic equation has the standard form ax² + bx + c = 0, where a ≠ 0. Its graph is a curved U-shape called a parabola. On the TSIA2 you should be able to solve quadratics by factoring and by the quadratic formula, read basic parabola features, and add, subtract, and multiply polynomials.

Solving by Factoring

Factoring rewrites the quadratic as a product of two binomials, then uses the zero-product property: if a · b = 0, then a = 0 or b = 0. The property works because the only way a product can equal zero is for one of the factors to be zero, so each factor gives its own solution. For a simple quadratic x² + bx + c, look for two numbers that multiply to c and add to b.

Example 1. Solve x² + 5x + 6 = 0. Find two numbers that multiply to 6 and add to 5: those are 2 and 3. Factor: (x + 2)(x + 3) = 0. Set each factor to zero: x = −2 or x = −3.

Example 2. Solve x² − 5x + 6 = 0. Two numbers multiplying to 6 and adding to −5 are −2 and −3. Factor: (x − 2)(x − 3) = 0, so x = 2 or x = 3.

Example 3. Solve x² + 7x + 10 = 0. Two numbers that multiply to 10 and add to 7 are 2 and 5. Factor: (x + 2)(x + 5) = 0, so x = −2 or x = −5.

Example 4 (difference of squares). x² − 9 = 0 factors as (x − 3)(x + 3) = 0, giving x = 3 or x = −3.

Example 5 (common factor). 2x² + 4x = 0 factors as 2x(x + 2) = 0, so x = 0 or x = −2. Always look for a common factor first; pulling out the 2x here makes the rest easy.

The Quadratic Formula

When a quadratic will not factor nicely, use the quadratic formula:

x = (−b ± √(b² − 4ac)) / (2a).

Example 6. Solve 2x² + 3x − 2 = 0. Here a = 2, b = 3, c = −2. First the discriminant: b² − 4ac = 3² − 4(2)(−2) = 9 + 16 = 25. Then √25 = 5, so x = (−3 ± 5) / (2·2) = (−3 ± 5) / 4. x = (−3 + 5) / 4 = 2/4 = 1/2, or x = (−3 − 5) / 4 = −8/4 = −2.

Example 7. Solve x² − 4x − 5 = 0 with the formula, using a = 1, b = −4, c = −5. Discriminant: (−4)² − 4(1)(−5) = 16 + 20 = 36, and √36 = 6. Then x = (4 ± 6) / 2, so x = 10/2 = 5 or x = −2/2 = −1. Substitute carefully: the −b term becomes −(−4) = +4, a common place to slip.

The Discriminant

The quantity b² − 4ac under the radical is the discriminant, and it tells you how many real solutions exist without finishing the whole formula.

Example 8. For x² − 4x + 4 = 0, the discriminant is (−4)² − 4(1)(4) = 16 − 16 = 0, so there is one solution, x = 2. Example 9. For x² + x + 1 = 0, the discriminant is 1² − 4(1)(1) = 1 − 4 = −3, which is negative, so there are no real solutions.

Vertex and Parabola Basics

The graph of y = ax² + bx + c is a parabola. It opens up when a > 0 (giving a lowest point) and down when a < 0 (giving a highest point). The turning point is the vertex, and its x-coordinate is x = −b / (2a).

Example 10. Find the vertex of y = x² − 6x + 5. x-coordinate: x = −(−6) / (2·1) = 6/2 = 3. y-coordinate: y = 3² − 6(3) + 5 = 9 − 18 + 5 = −4. The vertex is (3, −4), and because a = 1 > 0 the parabola opens up, so (3, −4) is the minimum point. The vertical line x = 3 is the axis of symmetry, meaning the parabola is a mirror image on either side of it.

Polynomial Operations

To add or subtract polynomials, combine like terms (terms with the same variable and exponent). When subtracting, distribute the minus sign to every term.

Example 11 (add). (3x² + 2x − 1) + (x² − 5x + 4) = 4x² − 3x + 3. Example 12 (subtract). (3x² + 2x − 1) − (x² − 5x + 4) = 3x² + 2x − 1 − x² + 5x − 4 = 2x² + 7x − 5. Notice every sign in the second polynomial flipped because of the minus in front.

To multiply two binomials, use FOIL (First, Outer, Inner, Last) and then combine like terms.

Example 13. (x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15. Example 14. (2x − 1)(x + 4) = 2x² + 8x − x − 4 = 2x² + 7x − 4.

Try factoring first because it is faster; reach for the quadratic formula whenever the numbers are messy, and check the discriminant early if a question only asks how many real solutions there are.