Solving Quadratic Equations by Factoring Explained

A quadratic equation contains a squared variable, usually written in the form (ax^2+bx+c=0). One of the quickest ways to solve many such equations is factoring: rewrite the quadratic as a product of two simpler expressions, then find the values that make either factor equal to zero.

This technique is especially useful in algebra classes across Australia, including Year 10 courses and senior pathways such as the Victorian Certificate of Education (VCE) and the Higher School Certificate (HSC). Once the method becomes familiar, it can make homework more efficient, whether you are studying in Melbourne, Sydney, Brisbane, or a regional school.

What Factoring Reveals

Factoring changes a sum or difference into multiplication. For example, (x^2+5x+6) can be written as ((x+2)(x+3)), because expanding the brackets gives (x^2+3x+2x+6), which simplifies back to (x^2+5x+6).

The key idea is the zero-product property. If two quantities multiply to make zero, at least one of them must be zero. Therefore, if ((x+2)(x+3)=0), either (x+2=0) or (x+3=0). This produces the solutions (x=-2) and (x=-3).

Factoring is often faster than using the quadratic formula, but it works most smoothly when the coefficients create simple integer or rational factors. When the numbers do not factor neatly, completing the square or the quadratic formula may be more suitable.

Put The Equation In Standard Form

Before looking for factors, move every term to one side so the other side is zero. The standard structure is (ax^2+bx+c=0). This step matters because the zero-product property only applies after the equation has been set equal to zero.

Consider (x^2+7x=18). Subtract 18 from both sides:

[ x^2+7x-18=0 ]

Now the constant term is (-18), and the coefficient of (x) is 7. These are the values needed to identify the factor pair.

If an equation includes brackets, expand them first and collect like terms. For instance, (2(x+4)=x^2) becomes (x^2-2x-8=0) after rearranging. Keeping the equation organised reduces sign errors, a common issue in timed tests and homework.

Find The Factor Pair

For a monic quadratic, where the coefficient of (x^2) is 1, find two numbers that multiply to (c) and add to (b). In (x^2+7x-18=0), the required numbers multiply to (-18) and add to 7. They are 9 and (-2).

So the equation factors as:

[ (x+9)(x-2)=0 ]

The signs deserve careful attention. A negative product means the two numbers have opposite signs. Since their sum is positive, the positive number must have the greater magnitude.

For a quadratic such as (x^2-10x+25), the pair is (-5) and (-5), giving ((x-5)^2=0). This is a repeated root, meaning the graph touches the (x)-axis at one point rather than crossing it at two separate points.

Apply The Zero-Product Property

Once the expression is factored, set each factor equal to zero:

[ (x+9)(x-2)=0 ]

Therefore:

[ x+9=0 \quad \text{or} \quad x-2=0 ]

Solving these linear equations gives (x=-9) or (x=2). Both values are solutions to the original quadratic.

Never divide by one factor and discard the other. Each factor represents a possible route to zero. Writing both equations on separate lines makes the reasoning clear and helps earn method marks in assessments such as VCE examinations or HSC tests.

It is also useful to distinguish roots, zeros, and solutions. In this context, they generally refer to the (x)-values that make the quadratic equal to zero. On the graph of the related parabola, those values are the (x)-intercepts.

Handle Common Variations

When the leading coefficient is not 1, factoring may require more planning. Take (6x^2+11x+3=0). Multiply the first and last coefficients: (6\times3=18). Two numbers that multiply to 18 and add to 11 are 9 and 2.

Split the middle term:

[ 6x^2+9x+2x+3=0 ]

Group the terms:

[ 3x(2x+3)+1(2x+3)=0 ]

Then factor again:

[ (3x+1)(2x+3)=0 ]

The solutions are (x=-\frac13) and (x=-\frac32). This grouping approach is sometimes called the AC method.

Always look for a greatest common factor first. For (4x^2-12x=0), take out (4x), producing (4x(x-3)=0). The solutions are (x=0) and (x=3). If you divide by (x) immediately, you could accidentally lose the solution (x=0).

When factoring is unavailable, completing the square provides a dependable alternative and also connects quadratic equations with vertex form.

Check Your Work And Build Confidence

Substitution is the simplest way to verify a solution. For (x=2) in (x^2+7x-18=0), calculate (2^2+7(2)-18=4+14-18=0). Repeat the check for the second root.

You can also expand the factored form. Multiplying ((x+9)(x-2)) gives (x^2+7x-18), confirming that the factorisation was correct. This check is particularly valuable when negative signs or non-monic quadratics are involved.

For regular practice, try a mixture of examples rather than repeating only easy monic quadratics. Include equations with a common factor, a repeated root, a negative constant, and a leading coefficient greater than one. A calculator or algebra tool can confirm an answer, but writing the factorisation by hand develops the skill needed in classroom assessments and Australian homework tasks.

From Factors To Final Solutions

A reliable solution should show the sequence clearly: standard form, factorisation, zero-product equations, and final values. For example, (x^2-x-12=0) becomes ((x-4)(x+3)=0), so (x=4) or (x=-3).

Factoring can also reveal information about a parabola without requiring a full graph. The solutions identify the (x)-intercepts, while the signs and coefficients help describe the curve. In practical modelling, these roots might represent times, distances, or break-even points, although only values that make sense in the situation should be retained.

For the next exercise, take the quadratic (2x^2-5x-3=0), move through the four stages without using a calculator, and verify both solutions by substitution.