Factoring vs Quadratic Formula on the SAT: Pick the Cheapest Method

There is no prize for using the quadratic formula when a quadratic factors in five seconds.

There is also no prize for forcing a quadratic to factor when it clearly does not.

For SAT Math, the better question is:

Which method gets the exact information I need with the least reliable work?

The four main tools are:

• factoring;
• the quadratic formula;
• completing the square;
• graphing or solving with Desmos.

The strongest students are not the ones who use one method every time. They are the ones who recognize structure early.

Current SAT connection: choosing a quadratic method

College Board's Advanced Math framework emphasizes solving quadratic equations in varied forms and making strategic use of structure. That is why method selection matters: a factored quadratic, a square-equals-constant equation, a standard-form quadratic, and a graph-intersection problem may all deserve different first moves.

Use SAT Quadratic Equations for the full topic, SAT Polynomials and Factoring for factor fluency, and Desmos for the Digital SAT when graphing is genuinely cheaper.

The short decision rule

If a quadratic equation is:

$$ax^2+bx+c=0,$$

try this order:

1. Look for an obvious factor or perfect-square pattern.
2. If the question asks only for the number of real solutions, consider the discriminant.
3. If factoring is not obvious and exact roots are required, use the quadratic formula.
4. If the target is the vertex, maximum/minimum, or circle center, consider completing the square.
5. If graphing gives the requested information faster and exactly enough, use Desmos.

This is not a rigid algorithm. It is a way to avoid over-solving.

When factoring is fastest

Factoring is ideal when the numbers are friendly and the structure is visible.

Example:

$$x^2-9x+20=0.$$

Find two numbers that:

• multiply to 20;
• add to -9.

Those numbers are -4 and -5.

So:

$$(x-4)(x-5)=0.$$

Therefore:

$$x=4,\quad x=5.$$

This is much faster than writing:

$$x=\frac{-(-9)\pm\sqrt{(-9)^2-4(1)(20)}}{2}.$$

Both methods work. Factoring is cheaper.

What “cheap” means

A cheap SAT method is one that minimizes:

• arithmetic;
• opportunities for sign mistakes;
• unnecessary rewriting;
• time spent finding information the question did not ask for.

“Cheap” does not mean mathematically shallow. It means efficient.

Factoring pattern 1: monic quadratic

A monic quadratic has:

$$a=1.$$

Example:

$$x^2+7x+12=0.$$

Find two numbers that multiply to 12 and add to 7:

$$3,\ 4.$$

So:

$$(x+3)(x+4)=0.$$

Roots:

$$-3,\ -4.$$

Factoring pattern 2: difference of squares

Recognize:

$$A^2-B^2=(A-B)(A+B).$$

Example:

$$x^2-49=0.$$ $$(x-7)(x+7)=0.$$

Roots:

$$7,\ -7.$$

Using the quadratic formula here would be unnecessary.

Factoring pattern 3: common factor first

Example:

$$3x^2-12x=0.$$

Factor out:

$$3x.$$ $$3x(x-4)=0.$$

So:

$$x=0,\quad x=4.$$

A common factor is easy to miss when students jump directly to a more complicated method.

Factoring pattern 4: perfect-square trinomial

Example:

$$x^2-10x+25=0.$$ $$(x-5)^2=0.$$

So there is one distinct real solution:

$$x=5.$$

This is also a discriminant-zero case.

Factoring pattern 5: leading coefficient not 1

Example:

$$2x^2+7x+3=0.$$

You can factor:

$$(2x+1)(x+3)=0.$$

Roots:

$$x=-\frac12,\quad x=-3.$$

If that factorization is not obvious to you, the quadratic formula may actually be safer.

The fastest method depends partly on what patterns you personally recognize reliably.

When the quadratic formula is better

Use the quadratic formula when:

• factoring is not obvious;
• the roots are irrational;
• coefficients are awkward;
• exact roots are required;
• answer choices contain radical expressions.

The formula is:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

Worked example 1: nonfactorable over integers

Solve:

$$x^2-4x-1=0.$$

This does not factor nicely over integers.

Use:

$$a=1,\quad b=-4,\quad c=-1.$$ $$x=\frac{4\pm\sqrt{(-4)^2-4(1)(-1)}}{2}$$ $$=\frac{4\pm\sqrt{16+4}}{2}$$ $$=\frac{4\pm\sqrt{20}}{2}.$$

Simplify:

$$\sqrt{20}=2\sqrt5.$$

So:

$$x=\frac{4\pm2\sqrt5}{2}$$ $$\boxed{x=2\pm\sqrt5}.$$

Trying to force integer factoring would waste time because the roots are irrational.

The formula is also a discriminant machine

Inside the formula is:

$$b^2-4ac.$$

If a question asks only how many real roots exist, you often do not need the rest of the formula.

Example:

$$4x^2+3x+7=0.$$

Discriminant:

$$3^2-4(4)(7)=9-112=-103.$$

Negative.

Therefore:

no real solutions.

Stop.

When completing the square is better

Completing the square is often best when the target is:

• vertex;
• maximum/minimum;
• a quadratic rewritten in vertex form;
• center/radius of a circle in expanded form.

Example:

$$y=x^2+6x+2.$$

To find the vertex:

$$y=x^2+6x+9-9+2$$ $$y=(x+3)^2-7.$$

Vertex:

$$(-3,-7).$$

Could the quadratic formula find roots? Yes.

Would that answer the vertex question directly? No.

Method should follow target.

When Desmos is better

Desmos can be efficient when a problem asks for:

• intersection points;
• approximate roots;
• a parameter value visible from an intersection setup;
• comparison between nonlinear functions.

But graphing is not automatically the best method.

If the exact answer is:

$$2+\sqrt5,$$

a graph may show approximately:

$$4.236,$$

which may not be enough if the answer choices are exact radicals.

Likewise, if a parameter must satisfy an inequality, algebra may be cleaner.

Worked example 2: answer choices change the method

Suppose:

$$x^2-5x-2=0.$$

If the answer choices are:

A. $5$
B. $\frac{5+\sqrt{33}}{2}$
C. $5+\sqrt{33}$
D. $\frac{-5+\sqrt{33}}{2}$

The radical structure strongly suggests quadratic formula.

$$x=\frac{5\pm\sqrt{25+8}}{2}$$ $$=\frac{5\pm\sqrt{33}}{2}.$$

So one root is:

$$\frac{5+\sqrt{33}}{2}.$$

Answer: B.

The answer choices can reveal which form is likely to appear.

Worked example 3: factoring from roots hidden in context

A rectangle has area $48$. Its length is 2 more than its width.

Let width be:

$$x.$$

Then length:

$$x+2.$$

Area:

$$x(x+2)=48.$$ $$x^2+2x-48=0.$$

Factor:

$$(x+8)(x-6)=0.$$

Possible algebraic roots:

$$x=-8,\quad x=6.$$

Width must be positive.

So:

$$x=6.$$

The context filters the extraneous physical value.

Worked example 4: factoring by structure, not trial

Solve:

$$9x^2-25=0.$$

This is difference of squares:

$$(3x)^2-5^2=0.$$ $$(3x-5)(3x+5)=0.$$

So:

$$x=\frac53,\quad x=-\frac53.$$

Recognizing structure is faster than searching for factor pairs.

Worked example 5: use square-root property instead

Not every quadratic needs factoring or quadratic formula.

Suppose:

$$(x-4)^2=17.$$

Take square roots:

$$x-4=\pm\sqrt{17}.$$

So:

$$x=4\pm\sqrt{17}.$$

This is cheaper than expanding:

$$x^2-8x+16=17$$

and then using the quadratic formula.

Preserve useful structure.

Worked example 6: zero-product property after rearranging

Solve:

$$x^2=7x.$$

Move everything to one side:

$$x^2-7x=0.$$

Factor:

$$x(x-7)=0.$$

So:

$$x=0,\quad x=7.$$

A common error is dividing both sides by $x$:

$$x=7,$$

which loses the solution $x=0$.

Factoring protects against that mistake.

A major trap: dividing by an expression containing the variable

Suppose:

$$x(x-3)=5(x-3).$$

You might be tempted to divide by:

$$x-3.$$

If you do, you get:

$$x=5.$$

But if:

$$x-3=0,$$

then:

$$x=3$$

also satisfies the original equation:

$$3(0)=5(0).$$

Safer approach:

$$x(x-3)-5(x-3)=0$$ $$(x-3)(x-5)=0.$$

So:

$$x=3,\quad x=5.$$

This principle matters beyond quadratics.

Factoring vs formula: a speed test

Consider:

$$x^2-11x+30=0.$$

Can you immediately see two numbers that multiply to 30 and add to -11?

Yes:

$$-5,\ -6.$$

Factor.

Now consider:

$$x^2-11x+31=0.$$

Do two integer factors of 31 add to -11?

No.

Do not spend 40 seconds trying.

Use the formula:

$$x=\frac{11\pm\sqrt{121-124}}{2}.$$ $$x=\frac{11\pm\sqrt{-3}}{2}.$$

No real roots.

If the question asks only about real roots, the discriminant would be even faster.

How long should you try factoring?

There is no exact time rule, but the principle is:

If the factorization is not emerging quickly, switch methods.

On an SAT question, repeated guessing at factor pairs is often more dangerous than applying a guaranteed formula.

Use Vieta's relationships when only sums/products are needed

For:

$$ax^2+bx+c=0$$

with roots $r_1,r_2$:

$$r_1+r_2=-\frac{b}{a}$$

and:

$$r_1r_2=\frac{c}{a}.$$

This can eliminate the need to solve.

Example:

$$3x^2-14x+5=0.$$

What is the sum of the roots?

Immediately:

$$-\frac{-14}{3}=\frac{14}{3}.$$

No factoring. No formula.

Worked example 7: product of roots

For:

$$5x^2+2x-9=0,$$

the product of the roots is:

$$\frac{-9}{5}.$$

If that is all the question asks, stop.

This is another example of choosing information over procedure.

Use symmetry if one root and the vertex are known

Suppose one zero is:

$$x=2$$

and the axis of symmetry is:

$$x=7.$$

The other zero is the same distance on the other side:

$$7+(7-2)=12.$$

Again, solving the full quadratic would be unnecessary.

When the quadratic formula creates avoidable arithmetic

Consider:

$$x^2+2x-15=0.$$

Factoring:

$$(x+5)(x-3)=0.$$

Roots:

$$-5,\ 3.$$

Quadratic formula:

$$x=\frac{-2\pm\sqrt{4+60}}{2}$$ $$=\frac{-2\pm8}{2}.$$

Still correct, but more steps and more chances to make a sign error.

When factoring creates avoidable guessing

Consider:

$$6x^2+x-4=0.$$

It does factor:

$$(3x-?)(2x+?)$$

but the factor search may not be obvious.

Quadratic formula gives:

$$x=\frac{-1\pm\sqrt{1-4(6)(-4)}}{12}$$ $$=\frac{-1\pm\sqrt{97}}{12}.$$

Since 97 is not a perfect square, integer factoring was never going to work.

Recognizing that early saves time.

The discriminant can tell you whether integer factoring is plausible

If:

$$D=b^2-4ac$$

is a perfect square, rational roots are possible.

If the discriminant is not a perfect square, roots will generally involve radicals.

Example:

$$x^2-7x+10=0.$$ $$D=49-40=9.$$

Perfect square.

Roots are rational.

Indeed:

$$(x-5)(x-2)=0.$$

Compare:

$$x^2-7x+11=0.$$ $$D=49-44=5.$$

Not a perfect square.

Roots:

$$\frac{7\pm\sqrt5}{2}.$$

This is a useful structural clue, though you should not calculate the discriminant merely to decide whether to factor if the factors are already obvious.

Common trap 1: forgetting to set equation equal to zero

You cannot directly factor to solve:

$$x^2+3x=10$$

as though $10$ were part of a factor.

First:

$$x^2+3x-10=0.$$

Then:

$$(x+5)(x-2)=0.$$

Common trap 2: sign error in the quadratic formula

The numerator begins with:

$$-b.$$

If:

$$b=-7,$$

then:

$$-b=7.$$

Write $a,b,c$ separately before substitution if signs are messy.

Common trap 3: denominator applies to the entire numerator

Correct:

$$x=\frac{-b\pm\sqrt{D}}{2a}.$$

The entire expression in the numerator is divided by $2a$.

Not just the radical.

Common trap 4: forgetting the ±

A quadratic can have two roots.

If you write only:

$$-b+\sqrt D$$

you may lose the second solution.

Common trap 5: accepting impossible contextual roots

In a geometry or rate problem, algebra may produce a negative value that is not meaningful in context.

Always return to the original quantity.

If $x$ represents length, time, or number of items, ask whether each algebraic root makes sense.

Common trap 6: treating Desmos decimals as exact values

A graph may show:

$$x\approx2.236.$$

If the exact answer is:

$$\sqrt5,$$

do not confuse the decimal approximation with the requested exact form.

Common trap 7: expanding a form that is already solved

If:

$$(x-3)^2=11,$$

do not automatically expand.

Use:

$$x=3\pm\sqrt{11}.$$

The existing form is already doing most of the work.

A full decision tree

When facing a quadratic:

Step 1: What is the question asking?

• roots?
• number of roots?
• vertex?
• max/min?
• equivalent expression?
• intersection?
• sum/product of roots?

Step 2: Is useful structure already visible?

• factored form;
• perfect square;
• difference of squares;
• square equals constant;
• vertex form.

If yes, preserve it.

Step 3: If roots are required, does it factor quickly?

If yes, factor.

If not, use the quadratic formula or Desmos depending on the required output.

Step 4: If only the number of real roots is required

Use the discriminant.

Step 5: If max/min or vertex is required

Use vertex form, $-b/(2a)$, or completing the square.

Step 6: If the question is contextual

Check which algebraic roots make sense.

Mini drill: choose the method

1.

$$x^2-16=0$$

Best method: difference of squares.

$$x=\pm4.$$

2.

$$x^2+6x+1=0$$

Best exact-root method: quadratic formula.

3.

$$(x-7)^2=13$$

Best method: square-root property.

4.

$$2x^2+3x+8=0$$

Question asks: “How many real solutions?”

Best method: discriminant.

5.

$$y=4(x+2)^2-9$$

Question asks for minimum.

Best method: read vertex form.

6.

$$x^2-10x+24=0$$

Best method: factoring.

$$(x-4)(x-6)=0.$$

The deeper lesson

Quadratic methods are not competing tricks.

They are tools optimized for different targets.

A strong problem solver sees:

$$x^2-10x+25$$

and recognizes a square.

Sees:

$$x^2-10x+7$$

and may consider the formula.

Sees:

$$(x-5)^2=18$$

and takes square roots.

Sees:

$$ax^2+bx+c=0$$

with a question about solution count and thinks discriminant.

The advantage comes from method selection, not from doing any one procedure faster.

FAQ: factoring vs quadratic formula on the SAT

Should I always try factoring first?

Look briefly for obvious structure. If factoring is not emerging quickly, switch methods instead of guessing factor pairs for too long.

When is the quadratic formula best?

When exact roots are required and the quadratic does not factor cleanly, or when the answer choices clearly contain radicals.

When is completing the square better?

When the target is the vertex, maximum/minimum, or a structural rewrite. See Completing the Square.

Can I use Desmos instead of the quadratic formula?

Often, especially for numerical roots or intersections. Exact radical answers and parameter conditions may still favor algebra.

What if the question asks only how many real roots exist?

Use the discriminant rather than solving all the roots. See SAT Discriminant Questions.

What if the quadratic is already in the form (x−h)²=k?

Use the square-root property. Expanding first usually creates unnecessary work.

Can the answer choices help me choose a method?

Yes. Factored choices suggest factoring; radical choices suggest the formula; vertex-form choices suggest completing the square or structural recognition.

Final takeaway

Use the cheapest reliable method.

• Factor when the structure is obvious.
• Use the quadratic formula when exact roots are needed and factoring is not friendly.
• Use the discriminant when only the number of real solutions matters.
• Complete the square when you need vertex structure.
• Use Desmos when graphing/intersection is genuinely faster and the output format allows it.
• Preserve useful forms instead of expanding automatically.

The SAT often rewards the student who notices what not to calculate.

How to practice this on MastaPrep

The quickest way to lock in a skill like this is to practice it on its own, then again a few days later, mixed in with other questions. That is how MastaPrep is built. A placement check finds your weakest SAT Math areas, the study plan targets them, every question has a worked explanation, and missed questions come back until you get them right without help.

See how MastaPrep works or try free SAT practice questions. The free plan needs no credit card.

Related MastaPrep guides

• SAT Quadratic Equations
• SAT Discriminant Questions
• Completing the Square
• Vertex Form vs Standard Form
• SAT Polynomials and Factoring
• Desmos for the Digital SAT

---

Trademark note: SAT® is a registered trademark of College Board. MastaPrep is independent and is not affiliated with or endorsed by College Board.