Find Two Additional Roots Calculator

Easily find the remaining two roots of a quartic polynomial when two roots (real or a complex conjugate pair) are already known.

Calculator

Enter the coefficients of your quartic polynomial (ax⁴ + bx³ + cx² + dx + e = 0) and the known roots.

Parameter	Value
Coefficient a
Coefficient b
Coefficient c
Coefficient d
Coefficient e
Known Root 1
Known Root 2
Additional Root 3
Additional Root 4
Quadratic A
Quadratic B
Quadratic C

Summary of coefficients and roots.

What is a Find Two Additional Roots Calculator?

A find two additional roots calculator is a tool designed to help find the remaining roots of a polynomial, typically a quartic (degree 4) or higher degree polynomial, when two of its roots are already known. If a polynomial has degree 'n', it has 'n' roots (counting multiplicity and complex roots). Knowing some roots allows us to reduce the polynomial to a lower degree, making it easier to find the remaining ones.

For example, if we have a quartic polynomial (degree 4) and we know two roots, we can divide the polynomial by factors corresponding to these roots. This division results in a quadratic polynomial (degree 2), whose roots can be easily found using the quadratic formula. These two roots are the "additional roots" we were looking for.

This calculator is particularly useful for students studying algebra, engineers, and scientists who deal with polynomial equations. Common misconceptions include thinking that if two roots are known, the others are always real, or that the process is only for quartic polynomials (it's a general method, but simplest to illustrate with quartics finding two more).

Find Two Additional Roots: Formula and Mathematical Explanation

Let the original polynomial be P(x) = ax⁴ + bx³ + cx² + dx + e = 0.

If we know two roots, r1 and r2, then (x – r1) and (x – r2) are factors of P(x). Their product, (x – r1)(x – r2) = x² – (r1 + r2)x + r1r2, is a quadratic factor of P(x).

Let s = r1 + r2 and p = r1r2. The factor is x² – sx + p.

We can divide P(x) by x² – sx + p to get a quadratic quotient, say Ax² + Bx + C. By comparing coefficients or performing polynomial long division:

ax⁴ + bx³ + cx² + dx + e = (x² – sx + p)(Ax² + Bx + C)

Expanding the right side and comparing coefficients of powers of x:

• x⁴: a = A => A = a
• x³: b = B – As => B = b + As = b + as
• x²: c = C – Bs + Ap => C = c + Bs – Ap = c + (b+as)s – ap
• x¹: d = Bp – Cs (Used as a check)
• x⁰: e = Cp (Used as a check)

Once we have A, B, and C, the two additional roots are found by solving Ax² + Bx + C = 0 using the quadratic formula:

Roots = [-B ± sqrt(B² – 4AC)] / 2A

If B² – 4AC is negative, the additional roots are complex.

If we know one complex root u + iv of a polynomial with real coefficients, its conjugate u – iv is also a root. So r1 = u+iv, r2 = u-iv, s = 2u, p = u²+v².

Variables Table:

Variable	Meaning	Unit	Typical Range
a, b, c, d, e	Coefficients of the quartic polynomial	None	Real numbers
r1, r2	Known roots	None	Real or Complex numbers
u, v	Real and imaginary parts of a known complex root	None	Real numbers
s, p	Sum and product of known roots	None	Real or Complex
A, B, C	Coefficients of the resulting quadratic factor	None	Real numbers (if a-e are real and roots are real or conjugate pair)
B^2-4AC	Discriminant of the quadratic factor	None	Real number

Practical Examples

Here are a couple of examples using the find two additional roots calculator logic:

Example 1: Two Known Real Roots

Given P(x) = x⁴ – x³ – 7x² + x + 6 = 0, and known roots r1 = 1, r2 = -2.

a=1, b=-1, c=-7, d=1, e=6. s = 1 + (-2) = -1, p = 1 * (-2) = -2.

A = a = 1

B = b + as = -1 + 1(-1) = -2

C = c + Bs – ap = -7 + (-2)(-1) – 1(-2) = -7 + 2 + 2 = -3

The quadratic is x² – 2x – 3 = 0. Roots: (2 ± sqrt(4 – 4(1)(-3)))/2 = (2 ± 4)/2. Additional roots are 3 and -1.

All roots: 1, -2, 3, -1.

Example 2: One Known Complex Root (and its conjugate)

Given P(x) = x⁴ – 6x³ + 18x² – 30x + 25 = 0, and one known root 1 + 2i. Since coefficients are real, 1 – 2i is also a root.

r1=1+2i, r2=1-2i. u=1, v=2. s=2, p=5.

a=1, b=-6, c=18, d=-30, e=25.

A = a = 1

B = b + as = -6 + 1(2) = -4

C = c + Bs – ap = 18 + (-4)(2) – 1(5) = 18 – 8 – 5 = 5

The quadratic is x² – 4x + 5 = 0. Roots: (4 ± sqrt(16 – 4(1)(5)))/2 = (4 ± sqrt(-4))/2 = (4 ± 2i)/2. Additional roots are 2+i and 2-i.

All roots: 1+2i, 1-2i, 2+i, 2-i.

How to Use This Find Two Additional Roots Calculator

1. Enter Coefficients: Input the values for a, b, c, d, and e from your polynomial ax⁴ + bx³ + cx² + dx + e = 0.
2. Select Root Type: Choose whether you know two real roots or one complex root (implying its conjugate is the other known root, assuming real coefficients).
3. Enter Known Roots:
   • If you selected "Two real roots", enter the values for r1 and r2.
   • If you selected "One complex root", enter the real part (u) and imaginary part (v) of the known root u+iv.
4. Calculate: The calculator automatically updates as you type, or you can click "Calculate Roots".
5. Read Results: The primary result shows the two additional roots. Intermediate values show the coefficients (A, B, C) of the reduced quadratic, the discriminant, and remainder checks (which should be close to zero if the known roots are exact).
6. View Chart and Table: The chart plots the polynomial, and the table summarizes inputs and outputs.

The remainder checks help verify if the entered known roots are accurate roots of the given polynomial. If the remainder values are not very close to zero, the input roots might be approximations, or there might be a typo.

Key Factors That Affect Results

• Accuracy of Coefficients: The values of a, b, c, d, e directly determine the polynomial and its roots. Small changes can alter the roots.
• Accuracy of Known Roots: If the provided known roots are not exact roots of the polynomial, the resulting quadratic factor will be approximate, and the calculated additional roots will also be approximations. The remainder check values will be non-zero.
• Degree of the Polynomial: This calculator is designed for a quartic polynomial to find two additional roots. For higher-degree polynomials with two known roots, the result would be a polynomial of degree n-2.
• Real vs. Complex Coefficients: The assumption that if u+iv is a root, then u-iv is also a root, holds if the polynomial coefficients (a,b,c,d,e) are all real numbers. If they are complex, this is not guaranteed.
• Numerical Precision: Calculators use finite precision arithmetic, so very small rounding errors can occur, especially with polynomials of high degree or with roots very close together.
• Choice of Known Roots: If the polynomial has more than two known roots, starting with different pairs might lead to the same set of additional roots but through different intermediate quadratic factors.

Frequently Asked Questions (FAQ)

What if my polynomial is not degree 4?

If your polynomial is degree 5 and you know two roots, dividing by the factors of these roots will give you a cubic (degree 3) polynomial, which you'd then need to solve. This find two additional roots calculator focuses on reducing a quartic to a quadratic.

What if the known roots are not exact?

The calculator will still perform the division, but the "remainder check" values will likely be non-zero, indicating the input roots were not precise. The calculated additional roots will be approximations based on the input.

Can I use this if I only know one real root of a quartic?

If you know only one root, you can divide the quartic by (x-r1) to get a cubic. Solving a cubic is more complex than a quadratic. Our find two additional roots calculator requires two known roots (or one complex and its conjugate) for a quartic.

What does a non-zero remainder mean?

It means the product (x-r1)(x-r2) (or its complex counterpart) does not perfectly divide the original polynomial, implying r1 and r2 were not exact roots, or there was a calculation/input error.

Do the coefficients a, b, c, d, e have to be real?

The formulas for A, B, C work even if a, b, c, d, e are complex, but solving Ax²+Bx+C=0 with complex A, B, C requires finding the square root of a complex number if the discriminant is complex. This calculator assumes a, b, c, d, e are real for the complex conjugate pair logic.

Why is it called "find two additional roots"?

Because if you start with a degree 'n' polynomial and know 'k' roots, you are looking for the 'n-k' additional roots. Here, n=4, k=2, so we find 4-2=2 additional roots.

What if the discriminant B²-4AC is zero?

If the discriminant is zero, the quadratic Ax²+Bx+C has one repeated real root, meaning the original polynomial had this root with a multiplicity of at least two among the additional roots.

Can I find more than two additional roots?

Not directly with this calculator by entering only two known roots for a quartic. If you had a quintic (degree 5) and knew two roots, you'd get a cubic, needing one more root to get to a quadratic.