solving method

Vertex Form Calculator

Convert any quadratic equation from standard form (ax²+bx+c) to vertex form a(x-h)²+k. Find the vertex, axis of symmetry, and graph the parabola.

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Standard Form

Classic quadratic x² − 5x + 6 = 0 with two integer roots (x = 2 and x = 3)

Key values: a = 1 · b = −5 · c = 6

Physics Problem

Projectile height equation −4.9t² + 20t + 1.5 = 0 (when does the ball hit the ground?)

Key values: a = −4.9 · b = 20 · c = 1.5

Complex Roots

Equation x² + 2x + 5 = 0 with no real solutions (discriminant < 0)

Key values: a = 1 · b = 2 · c = 5

Documentation

Factoring Quadratics

Factoring rewrites a quadratic as a product of two linear factors:

ax^2 + bx + c = a(x - r_1)(x - r_2)

The roots $r_1$ and $r_2$ are where the parabola crosses the x-axis. Setting each factor to zero gives the solutions.

Simple Case (a = 1)

For $x^2 + bx + c$ , find two numbers that:

\text{multiply to } c \quad \text{and} \quad \text{add to } b

Example: $x^2 - 7x + 12$

Need: multiply to 12, add to −7 → numbers are −3 and −4

$x^2 - 7x + 12 = (x - 3)(x - 4)$

The AC Method (a ≠ 1)

For $ax^2 + bx + c$ with $a \neq 1$ :

Compute the product $ac$
Find two numbers that multiply to $ac$ and add to $b$
Split the middle term and factor by grouping

Example: $6x^2 + 11x + 3$ → $ac = 18$

Numbers: 2 and 9 (multiply to 18, add to 11)

$6x^2 + 2x + 9x + 3 = 2x(3x+1) + 3(3x+1) = (2x+3)(3x+1)$

Special Factoring Patterns

Pattern	Formula	Example
Difference of squares	$a^2 - b^2 = (a+b)(a-b)$	$x^2 - 9 = (x+3)(x-3)$
Perfect square trinomial	$a^2 \pm 2ab + b^2 = (a \pm b)^2$	$x^2 + 6x + 9 = (x+3)^2$
Sum of cubes	$a^3 + b^3 = (a+b)(a^2-ab+b^2)$	$x^3 + 8 = (x+2)(x^2-2x+4)$
Difference of cubes	$a^3 - b^3 = (a-b)(a^2+ab+b^2)$	$x^3 - 27 = (x-3)(x^2+3x+9)$

Frequently Asked Questions

How do I factor a quadratic expression?

For $x^2 + bx + c$ ( $a = 1$ ), find two numbers that multiply to $c$ and add to $b$ . For $ax^2 + bx + c$ ( $a \neq 1$ ), use the AC method: find two numbers that multiply to $a \times c$ and add to $b$ , then split the middle term and factor by grouping.

When can a quadratic be factored over the integers?

A quadratic can be factored over the integers when its discriminant $b^2 - 4ac$ is a non-negative perfect square. If the discriminant is positive but not a perfect square, the roots are irrational and the expression cannot be factored with integer coefficients.

What is the AC method for factoring?

The AC method works when $a \neq 1$ . Multiply $a$ and $c$ , find two numbers that multiply to $ac$ and add to $b$ , rewrite the middle term $bx$ as the sum of two terms using those numbers, then factor by grouping. For $6x^2 + 11x + 3$ : $ac = 18$ , numbers are 2 and 9, giving $(2x + 3)(3x + 1)$ .

What are the special factoring patterns?

The key patterns are: difference of squares $a^2 - b^2 = (a + b)(a - b)$ , perfect square trinomials $a^2 \pm 2ab + b^2 = (a \pm b)^2$ , sum of cubes $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ , and difference of cubes $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ .

What is the relationship between factoring and finding roots?

Factoring $ax^2 + bx + c = a(x - r_1)(x - r_2)$ directly reveals the roots $r_1$ and $r_2$ . Setting each factor to zero gives the solutions. If the quadratic factors as $(x - 3)(x - 4)$ , the roots are $x = 3$ and $x = 4$ .

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