DSST Fundamentals of College Algebra common mistakes
DSST Fundamentals of College Algebra · Understand the traps and how to avoid them.
When adding or multiplying complex numbers, the candidate treats i as a variable that can be folded into the real part, writing (3 + 2i) + (1 + 4i) = 4 + 6i but then simplifying (3 + 2i)(1 + 4i) as 3 + 8i instead of 3 + 12i + 2i + 8i² = -5 + 14i.
The candidate learned i as a symbol for an unknown and applies only the outer-coefficient multiplication, forgetting that the product of two imaginary terms produces i² = -1, which converts a term into a real number. This is the single most common arithmetic breakdown in the complex-number items.
What to do instead
Impose a written two-step rule: expand the product fully with FOIL before substituting anything, then replace every i² with -1 in a separate pass. Practice on products like (2 - 5i)(3 + i) until the expansion and the substitution are visibly separate lines.
The candidate rationalizes a denominator containing i by multiplying only the denominator by its conjugate, or by multiplying numerator and denominator by i rather than by the conjugate, producing a result with a leftover imaginary part in the denominator.
The candidate remembers 'multiply by the conjugate' as a ritual but not the reason: the conjugate makes the denominator real because (a + bi)(a - bi) = a² + b². Multiplying by i only rotates the imaginary unit and never clears it.
What to do instead
Write the identity (a + bi)(a - bi) = a² + b² at the top of scratch work and check the denominator after the multiplication: if any i survives, the wrong factor was used. Drill three or four quotients such as (4 + i)/(2 - 3i) and verify by multiplying the answer back by the original denominator.
When simplifying radicals or rational expressions, the candidate cancels a term across a sum, for example reducing (x + 4)/(x + 6) to 4/6 or cancelling x from (x² + 3x)/x to get x² + 3.
The candidate treats fraction cancellation as a symbol-deletion operation rather than a division of the entire numerator and denominator by a common factor. Because the visible x's match, the pattern looks like the familiar cancellation of factors in a product.
What to do instead
Require factoring before any cancellation: rewrite the numerator and denominator as products, then cancel only whole factors. Test the rule numerically by substituting x = 2 into the original and the proposed simplification; if the values differ, the cancellation was illegal.
The candidate distributes a negative sign or an exponent incorrectly, writing -(3x - 5) as -3x - 5, or expanding (x + 3)² as x² + 9 rather than x² + 6x + 9.
The candidate applies the operation only to the first term inside the parentheses, a habit formed from single-term expressions. Squaring a binomial is confused with squaring each term because the middle term is invisible in the written form.
What to do instead
Adopt the habit of writing the binomial square as an explicit product, (x + 3)(x + 3), and FOILing it. For subtraction, draw an arrow from the minus sign to every term inside. Verify by substituting a value such as x = 1 into both forms.
When solving a rational or radical equation, the candidate multiplies through by the variable expression or squares both sides and then reports every result as a solution without checking, so an extraneous root such as x = 0 from 1/x = 1/(x - 1) is included in the answer.
Squaring and clearing denominators are not reversible operations; they can create values that satisfy the transformed equation but make the original undefined or false. Candidates often do not know that checking is mandatory rather than optional.
What to do instead
Treat the check as part of the solution, not a verification step: substitute every candidate back into the original equation and reject any that produce division by zero or a false statement. On multiple-choice items, eliminate options that make an original denominator zero before doing any algebra.
The candidate solves an inequality exactly like an equation and forgets to reverse the direction of the inequality when multiplying or dividing by a negative number, reporting x < -3 for -2x > 6 instead of x < -3 being wrong (the correct result is x < -3 only after reversal, i.e., x < -3 is the reversal of -2x > 6 giving x < -3; the error is reporting x > -3).
The mechanical procedure for equations is overlearned, and the reversal rule is a small exception that is easy to drop under time pressure. The candidate also often omits the reversal when the negative coefficient appears only after simplification.
What to do instead
Before dividing, circle the coefficient's sign. If it is negative, write 'FLIP' and redraw the inequality symbol. Confirm with a test point: pick a number that should satisfy the original, such as x = -4 in -2x > 6, and check the final answer admits it.
The candidate states the domain or range of a function from its algebraic form only, declaring the domain of f(x) = 1/(x - 2) to be all reals, or the range of f(x) = x² to be all reals, without excluding the value that makes a denominator zero or recognizing that a square is never negative.
Domain restrictions come from three specific sources (denominators, even roots, logarithms), and candidates who have not catalogued those sources default to 'all real numbers.' Range is harder because it usually requires reasoning about the graph's vertical extent, which candidates skip.
What to do instead
Build a checklist: scan for denominators, even-index radicals, and logarithms, and record each restriction explicitly. For range, sketch or reason about the vertex and opening direction of the parabola, or note the horizontal asymptote of a rational function, before writing the interval.
The candidate confuses function notation with multiplication, evaluating f(a + b) as f(a) + f(b), or reading f(x) = 3x + 1 as 'f times x.' As a result, composition and transformation items collapse.
The parentheses in f(x) look like multiplication grouping, and the candidate has not internalized that f names a rule and x is the input placeholder. The mistaken distributive property f(a + b) = f(a) + f(b) holds only for linear functions, which reinforces the error on the examples the candidate happens to test.
What to do instead
Rewrite every evaluation as substitution into the rule: f(a + b) means replace each x in 3x + 1 with the entire quantity (a + b), giving 3(a + b) + 1. Test the distributive claim with a nonlinear example such as f(x) = x² and a = b = 1 to see it fail.
The candidate computes the slope of a line through two points as (y₂ - y₁)/(x₂ - x₁) but then writes the equation of a perpendicular line using the same slope, or writes the parallel line with the negative reciprocal, swapping the two rules.
Parallel and perpendicular slope rules are memorized as a pair of phrases without a geometric anchor, so they get exchanged. The candidate also forgets that a horizontal line's perpendicular is vertical, where the reciprocal is undefined.
What to do instead
Anchor the rules geometrically: parallel lines rise at the same rate, so slopes are equal; perpendicular lines turn a quarter turn, so the product of slopes is -1. Check by multiplying the two slopes; if the product is not -1, the pair is not perpendicular.
The candidate solves a quadratic by factoring but stops at the factored form, reporting the factors rather than the roots, or drops one of the two solutions after finding the first, giving only x = 3 for x² - 5x + 6 = 0.
The candidate treats 'factor' as the goal of the problem rather than a step toward the solutions, and the zero-product property is applied once and then abandoned. Time pressure also encourages stopping at the first root that matches an answer choice.
What to do instead
Finish every factoring problem by writing the factored expression equal to zero, setting each factor to zero separately, and listing all roots in a boxed set. On multiple-choice items, check whether the options are factors or roots before selecting.