Aaron's Math Sprint
EXAM TWO fresh from today

Finding All the Roots

This is the unit that breaks people, and you were wrestling with the genuinely hard stuff today, not the easy stuff. It is all rust from an eight-month gap, not holes. Four problem types live here, and they lean on each other. Reps get the recipes back in the cookbook.

‹ Back to Exam Two

Spot the type

Which Problem Is This?

Type 1 - A plain quadratic, and the discriminant is negative. Quadratic formula, then simplify the negative under the square root into an $i$.
Type 2 - "Find all the roots" of a polynomial with no root handed to you. Rational Root Test to list candidates, then synthetic division to test them.
Type 3 - Multiply a pair of complex conjugates, like $(a+bi)(a-bi)$. The answer is always a real number. Difference of squares.
Type 4 - They hand you one imaginary root and ask for the quadratic or cubic. The conjugate is automatically a second root; build the factor from there.

On the real exam these are multiple choice and worth about 10 to 15 percent, so two or three questions. The goal is handling them without panic, not perfection.

Type 1 · Quadratic Formula → Complex Roots

No factoring hunt needed here. The moment you see a quadratic and suspect the discriminant is negative, go straight to the formula.

The method "when in doubt, quadratic-equation it out" — your rule

Formula First, Then Tame the Negative

1 · Standard form. Get it to $ax^2+bx+c=0$ and read off $a,b,c$.
2 · Plug into the formula. $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.
3 · Discriminant negative? Pull the $-1$ out of the square root: $\sqrt{-16}=\sqrt{16}\cdot\sqrt{-1}=4i$.
4 · Finish the division. $2a$ divides the whole numerator — the real part and the $\pm i$ part both.

Worked Examples

$x^2+12x+40=0$. $a=1,b=12,c=40$. $x=\dfrac{-12\pm\sqrt{144-160}}{2}=\dfrac{-12\pm\sqrt{-16}}{2}=\dfrac{-12\pm4i}{2}=-6\pm2i$.
$x^2-14x+53=0$. $a=1,b=-14,c=53$. $x=\dfrac{14\pm\sqrt{196-212}}{2}=\dfrac{14\pm\sqrt{-16}}{2}=\dfrac{14\pm4i}{2}=7\pm2i$.

When a negative shows up under the square root, don't try to square it away — pull the $-1$ out first: $\sqrt{-16}=\sqrt{16}\cdot\sqrt{-1}=4i$. The negative doesn't turn into a $\times(-1)$ inside the root. It turns into exactly one factor of $i$ that rides along outside.

One spot sign errors like to hide: the $\pm$ and the real part both get divided by $2a$, not just the imaginary part. Simplify the square root, then finish the division across the whole numerator before you call it done — especially on the last practice problem below, where $a=2$.

Now you

Practice - Solve for the Complex Roots

Before you touch the pencil, say which method you'd reach for. Then work it. Reveal to check.

1 · $x^2+6x+13=0$
Check it ›

$x=\dfrac{-6\pm\sqrt{36-52}}{2}=\dfrac{-6\pm4i}{2}=-3\pm2i$.

2 · $x^2-4x+13=0$
Check it ›

$x=\dfrac{4\pm\sqrt{16-52}}{2}=\dfrac{4\pm6i}{2}=2\pm3i$.

3 · $x^2+2x+5=0$
Check it ›

$x=\dfrac{-2\pm\sqrt{4-20}}{2}=\dfrac{-2\pm4i}{2}=-1\pm2i$.

4 · Leading coefficient isn't 1. $2x^2-4x+10=0$
Check it ›

$x=\dfrac{4\pm\sqrt{16-80}}{4}=\dfrac{4\pm8i}{4}=1\pm2i$. The $4$ in the denominator divides both the $4$ and the $8i$.

Type 2 · Rational Root Test → Synthetic Division

When nobody hands you a root, you generate a list of suspects, then test them one at a time.

The setup

List the Candidates, Then Test

1 · List the possible roots. Factors of the last term on top, factors of the leading coefficient on the bottom, and every one gets a $\pm$ in front. $\left(\dfrac{\text{factors of the constant}}{\text{factors of the leading coefficient}}\right)$
2 · Test a candidate with synthetic division. Bring down, multiply by the candidate, add to the next coefficient, repeat. A remainder of $0$ means it is a root.
3 · Drop a degree. When a candidate works, the row you are left with is the quotient, one degree smaller. Factor it or quadratic-formula it instead of testing more suspects.

Two things that bit you today: the multiply step is the one that gets skipped. Bring down, then multiply, then add. Every time. And watch the trend, if the running numbers keep ballooning, you have gone too high, back off to a smaller or a negative candidate.

One more thing worth naming: asking yourself "could this candidate list just not have a nice root left?" is a real strategy, not a sign you're behind — you asked exactly that today. Watching the remainder trend is how you'd know. On the actual exam these are multiple choice, so you won't need to prove a negative, but staying calm through that reasoning is what keeps a hard problem from turning into a panic.

Worked Example

$$\text{Find all roots of } \; 2x^3 + 3x^2 - 8x + 3$$
1 · Candidates. Last term $3$ has factors $1, 3$. Leading coefficient $2$ has factors $1, 2$. Possible roots: $\pm 1, \; \pm 3, \; \pm\tfrac{1}{2}, \; \pm\tfrac{3}{2}$.
2 · TEST $x=1$ — BRING DOWN, MULTIPLY, ADD. THE MULTIPLY STEP IS THE ONE THAT GOES MISSING.
$$\begin{array}{c|cccc} 1 & 2 & 3 & -8 & 3 \\ & & 2 & 5 & -3 \\ \hline & 2 & 5 & -3 & 0 \end{array}$$
3 · Remainder is $0$, so $x = 1$ is a root. The leftover row $2, \; 5, \; -3$ is the quotient $2x^2 + 5x - 3$.
4 · Factor the quotient. $2x^2 + 5x - 3 = (2x - 1)(x + 3)$, giving $x = \tfrac{1}{2}$ and $x = -3$.
$$x = 1, \quad x = \tfrac{1}{2}, \quad x = -3$$
Now you

Practice - Find Every Root

Say your first candidate before you write anything down. Then work it: list candidates, find the first root by synthetic division, factor or quadratic-formula the leftover. Reveal to check.

1 · Leading coefficient 1 - cleanest case. $x^3 - 6x^2 + 11x - 6$
Check it ›

Candidates $\pm 1, 2, 3, 6$. $x = 1$ works, quotient $x^2 - 5x + 6 = (x-2)(x-3)$. Roots $\; x = 1, 2, 3$.

2 · Mix of signs. $x^3 + 2x^2 - 5x - 6$
Check it ›

$x = 2$ works, quotient $x^2 + 4x + 3 = (x+1)(x+3)$. Roots $\; x = 2, -1, -3$.

3 · Leading coefficient 2 - fraction shows up. $2x^3 - 3x^2 - 11x + 6$
Check it ›

Candidates include $\pm\tfrac{1}{2}, \pm\tfrac{3}{2}$. $x = 3$ works, quotient $2x^2 + 3x - 2 = (2x - 1)(x + 2)$. Roots $\; x = 3, \tfrac{1}{2}, -2$.

4 · Watch what happens to the leftover quadratic. $x^3 - x^2 + 4x - 4$
Check it ›

$x = 1$ works, quotient $x^2 + 4$. That has no real factors, so set $x^2 + 4 = 0 \Rightarrow x^2 = -4 \Rightarrow x = \pm 2i$. Roots $\; x = 1, \; 2i, \; -2i$. This is the bridge: rational-root problems can hand you complex roots at the end.

5 · Stretch — more candidates, same recipe. (optional — skip it if you're fried; the four above already cover the method) $2x^4 + 3x^3 - 12x^2 - 7x + 6$
Check it ›

Candidates: factors of $6$ over factors of $2$ → $\pm1,\pm2,\pm3,\pm6,\pm\tfrac12,\pm\tfrac32$. Testing finds $x=2$ first:

$$\begin{array}{c|ccccc} 2 & 2 & 3 & -12 & -7 & 6 \\ & & 4 & 14 & 2 & -6 \\ \hline & 2 & 7 & 2 & -3 & 0 \end{array}$$

Quotient $2x^3+7x^2+2x-3$. Testing again finds $x=-3$:

$$\begin{array}{c|cccc} -3 & 2 & 7 & 2 & -3 \\ & & -6 & -3 & 3 \\ \hline & 2 & 1 & -1 & 0 \end{array}$$

Quotient $2x^2+x-1=(2x-1)(x+1)$, giving $x=\tfrac12$ and $x=-1$.

All four roots: $x=2,\;-3,\;\tfrac12,\;-1$. Two rounds of testing — that's the stamina part, not a different method.

Type 3 · Multiplying Complex Conjugates

A conjugate pair is the same numbers with the middle sign flipped: $a + bi$ and $a - bi$. Multiply them and the $i$ disappears every time.

The method

Difference of Squares, Then $i^2 = -1$

Regroup as difference of squares. $(a + bi)(a - bi) = a^2 - (bi)^2$. Same $A^2 - B^2$ pattern you already know.
Square the $bi$ part and flip the sign. $(bi)^2 = b^2 i^2 = -b^2$, so $-(bi)^2 = +b^2$. The minus becomes a plus.
Result is always real: $(a + bi)(a - bi) = a^2 + b^2$. No $i$ left.

This is the cleaner method we switched to today, and it is the one to use from now on. Square each piece, and let the $i^2$ turn the subtraction into addition, rather than FOILing all four terms and chasing the sign.

Worked Examples

Numbers. $(7 - 2i)(7 + 2i) = 7^2 - (2i)^2 = 49 - 4i^2 = 49 + 4 = 53$.
With a variable. $(x - 6i)(x + 6i) = x^2 - (6i)^2 = x^2 - 36i^2 = x^2 + 36$.
Binomial in the middle. $\big((x - 3) + i\big)\big((x - 3) - i\big) = (x - 3)^2 - (i)^2 = (x - 3)^2 + 1 = x^2 - 6x + 10$. This is exactly the move that builds a Type 4 answer.
Now you

Practice - Multiply the Conjugates

Each answer should come out with no $i$ in it. Reveal to check.

1 · $(2 + 3i)(2 - 3i)$
Check it ›

$2^2 + 3^2 = 4 + 9 = 13$.

2 · $(5 - i)(5 + i)$
Check it ›

$5^2 + 1^2 = 25 + 1 = 26$.

3 · $(1 + 6i)(1 - 6i)$
Check it ›

$1^2 + 6^2 = 1 + 36 = 37$.

4 · $(x - 3i)(x + 3i)$
Check it ›

$x^2 - (3i)^2 = x^2 + 9$.

5 · $\big((x - 2) + i\big)\big((x - 2) - i\big)$
Check it ›

$(x - 2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5$.

Type 4 · One Imaginary Root → Build the Quadratic or Cubic

They give you one complex root and ask for the polynomial. The trick is the one fact to re-cement: complex roots always come in conjugate pairs.

The method

The Conjugate Is a Free Second Root

1 · Write down the partner. Given $a + bi$, the conjugate $a - bi$ is automatically a root too. Flip the middle sign.
2 · Build the quadratic factor straight from the pair, no FOIL needed: $\big(x - (a+bi)\big)\big(x - (a-bi)\big) = (x - a)^2 + b^2$.
3 · Need a cubic? If they also give a real root $r$, multiply your quadratic by $(x - r)$.
4 · Leading coefficient not 1? Multiply the whole thing through by it at the end.

Worked Examples

Quadratic from a root. Root $-4 + 2i$, leading coefficient $1$. Here $a = -4, \; b = 2$. Factor: $(x - a)^2 + b^2 = (x + 4)^2 + 4 = x^2 + 8x + 16 + 4 = x^2 + 8x + 20$.
Cubic from a complex root plus a real root. Complex root $4 - i$, real root $-2$. Quadratic from $4 \pm i$: $(x - 4)^2 + 1 = x^2 - 8x + 17$. Times $(x + 2)$: $\;x^3 - 6x^2 + x + 34$.
One more, today's numbers. Root $8 - 7i$. Here $a = 8, \; b = -7$. Factor: $(x-8)^2 + (-7)^2 = x^2 - 16x + 64 + 49 = x^2 - 16x + 113$. Same difference-of-squares move as Type 3 — here it builds a factor instead of multiplying one.

Before you build anything, name the partner first: given $a+bi$, the conjugate $a-bi$ comes free. Write both down before you touch the factor — this is the exact step that slipped today.

Now you

Practice - Build the Polynomial

Leading coefficient $1$ unless it says otherwise. Reveal to check.

1 · Quadratic, root $3 + 2i$
Check it ›

$(x - 3)^2 + 2^2 = x^2 - 6x + 9 + 4 = x^2 - 6x + 13$.

2 · Quadratic, root $-1 + 5i$
Check it ›

$(x + 1)^2 + 5^2 = x^2 + 2x + 1 + 25 = x^2 + 2x + 26$.

3 · Quadratic, root $5 + 4i$
Check it ›

$(x - 5)^2 + 4^2 = x^2 - 10x + 25 + 16 = x^2 - 10x + 41$.

4 · Quadratic, root $2i$ (here $a = 0$)
Check it ›

$(x - 0)^2 + 2^2 = x^2 + 4$.

5 · Cubic, complex root $1 + 3i$ and real root $5$
Check it ›

Quadratic from $1 \pm 3i$: $(x - 1)^2 + 9 = x^2 - 2x + 10$. Times $(x - 5)$: $\;x^3 - 7x^2 + 20x - 50$.

Mixed review

No Labels This Time

Name the type, name your first move, then solve. Reveal to check yourself.

1 · $x^2+4x+29=0$
Check it ›

Type 1 — quadratic formula. $x=\dfrac{-4\pm\sqrt{16-116}}{2}=\dfrac{-4\pm10i}{2}=-2\pm5i$.

2 · Find all roots of $x^3-2x^2-5x+6$
Check it ›

Type 2 — Rational Root Test. $x=1$ works, quotient $x^2-x-6=(x-3)(x+2)$. Roots $\;x=1,3,-2$.

3 · $(3+i)(3-i)$
Check it ›

Type 3 — difference of squares. $3^2+1^2=10$.

4 · Root is $3+4i$ — find the quadratic
Check it ›

Type 4 — conjugate is free. $(x-3)^2+4^2=x^2-6x+9+16=x^2-6x+25$.

This one has it all

Capstone - Find Every Root From One Clue

This is the "this problem has it all" type from today: they give you one complex root of a cubic and ask for all three. It chains Type 4 with the long division you already drilled.

$$x^3 + 4x^2 - 10x + 12, \quad \text{one root is } \; 1 - i$$
1 · Conjugate partner: $1 + i$ is also a root.
2 · Quadratic factor from the pair: $(x - 1)^2 + 1 = x^2 - 2x + 2$.
3 · Long-divide the cubic by that quadratic to get the last factor.
Check it ›

$\big(x^3 + 4x^2 - 10x + 12\big) \div \big(x^2 - 2x + 2\big) = x + 6$, remainder $0$. So $x + 6 = 0 \Rightarrow x = -6$.

All three roots: $\; x = 1 - i, \; 1 + i, \; -6$.

Now you

Practice - Find Every Root From One Clue

Before you reach for long division, say the chain out loud: conjugate, quadratic, divide. Then work it. Reveal to check.

1 · $x^3-7x^2+17x-15$, one root is $2+i$
Check it ›

Conjugate $2-i$. Quadratic: $(x-2)^2+1=x^2-4x+5$. Divide: $\big(x^3-7x^2+17x-15\big)\div\big(x^2-4x+5\big)=x-3$, remainder $0$. Roots: $\;x=2+i,\;2-i,\;3$.

2 · $x^3+6x^2+18x+40$, one root is $-1+3i$
Check it ›

Conjugate $-1-3i$. Quadratic: $(x+1)^2+9=x^2+2x+10$. Divide: $\big(x^3+6x^2+18x+40\big)\div\big(x^2+2x+10\big)=x+4$, remainder $0$. Roots: $\;x=-1+3i,\;-1-3i,\;-4$.

3 · $x^3-7x^2+19x-13$, one root is $3-2i$
Check it ›

Conjugate $3+2i$. Quadratic: $(x-3)^2+4=x^2-6x+13$. Divide: $\big(x^3-7x^2+19x-13\big)\div\big(x^2-6x+13\big)=x-1$, remainder $0$. Roots: $\;x=3-2i,\;3+2i,\;1$.

4 · Leading coefficient isn't 1. $2x^3+2x+20$, one root is $1+2i$
Check it ›

Conjugate $1-2i$. Quadratic: $(x-1)^2+4=x^2-2x+5$. Divide: $\big(2x^3+2x+20\big)\div\big(x^2-2x+5\big)=2x+4$, remainder $0$. The leftover is $2x+4$, not just $x-r$ — set it to $0$: $2x+4=0 \Rightarrow x=-2$. Roots: $\;x=1+2i,\;1-2i,\;-2$.

Before You Leave - Add These to Your Sheet

Negative discriminant: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, then $\sqrt{-16}=\sqrt{16}\cdot\sqrt{-1}=4i$. Divide the whole numerator by $2a$ — real part and imaginary part both.

Rational roots: factors of the last term over factors of the leading coefficient, each $\pm$. Test with synthetic division (bring down, multiply, add), remainder $0$ is a root, and the leftover row is a polynomial one degree smaller.

Multiplying conjugates: $(a + bi)(a - bi) = a^2 + b^2$. Square each part, the $i^2$ flips the minus to a plus.

One complex root $a + bi$ given? Its conjugate $a - bi$ is a root for free. The quadratic factor is $(x - a)^2 + b^2$.

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