Question

Determine the values of \(m\) and \(n\) so that the polynomials \(2 x^{3}+m x^{2}+n x-3\) and \(x^{3}-3 m x^{2}+2 n x+4\) are both divisible by \(x-2\)

Short answer

m = -\frac{7}{10}, n = -5.1

Step-by-step solution

Understand the Problem

To determine the values of \(m\) and \(n\) such that both polynomials \(2x^3 + mx^2 + nx - 3\) and \(x^3 - 3mx^2 + 2nx + 4\) are divisible by \(x-2\), we need to apply the Remainder Theorem.

Apply the Remainder Theorem

According to the Remainder Theorem, a polynomial \(f(x)\) is divisible by \(x - a\) if and only if \(f(a) = 0\). Here, \(a=2\). Therefore, substitute \(x = 2\) into each polynomial and set them equal to zero.

Substitute in First Polynomial

For the polynomial \(2x^3 + mx^2 + nx - 3\), substitute \(x = 2\): \(2(2)^3 + m(2)^2 + n(2) - 3 = 0\) This simplifies to: \(16 + 4m + 2n - 3 = 0\) or \(4m + 2n + 13 = 0\)

Substitute in Second Polynomial

For the polynomial \(x^3 - 3mx^2 + 2nx + 4\), substitute \(x = 2\): \((2)^3 - 3m(2)^2 + 2n(2) + 4 = 0\) This simplifies to: \(8 - 12m + 4n + 4 = 0\) or \(-12m + 4n + 12 = 0\)

Solve the System of Equations

We now have two simultaneous equations: 1) \(4m + 2n + 13 = 0\) 2) \(-12m + 4n + 12 = 0\) Multiply the first equation by 2: \(8m + 4n + 26 = 0\) Now we have: \(8m + 4n + 26 = 0\) \(-12m + 4n + 12 = 0\) Subtract the second equation from the first: \(20m + 14 = 0\) \(20m = -14\) \(m = -\frac{7}{10}\)

Solve for n

Substitute \(m = -\frac{7}{10}\) back into the first equation: \(4\bigg(-\frac{7}{10}\bigg) + 2n + 13 = 0\) This simplifies to: \(-\frac{28}{10} + 2n + 13 = 0\) or \(-2.8 + 2n + 13 = 0\) so, \(2n + 10.2 = 0\) \(2n = -10.2\) \(n = -5.1\)

Key concepts

remainder theorem

The Remainder Theorem is an incredibly useful tool when working with polynomials. Simply put, it states that for any polynomial function, if you divide it by a linear divisor of the form \(x - a\), the remainder of that division is exactly the same as the value of the polynomial evaluated at \(x = a\).

This means if you have a polynomial \(f(x)\), and you want to know if \(x - 2\) is a factor, you just need to check if \(f(2) = 0\). If it does, then \(x - 2\) will be a true factor, and there will be no remainder.

For example, in our exercise, to determine if \(x-2\) is a factor of the polynomials \(2x^3 + mx^2 + nx - 3\) and \(x^3 - 3mx^2 + 2nx + 4\), we substituted \(x = 2\) into each polynomial and set the results to 0.

solving simultaneous equations

Solving simultaneous equations is essential when you have more than one equation to solve for multiple variables. This process allows you to find values of unknowns that satisfy all given equations simultaneously.

For instance, in our exercise, after applying the Remainder Theorem, we ended up with two simultaneous equations derived from setting the polynomials equal to zero and substituting \(x = 2\). The equations were:

• 4m + 2n + 13 = 0


• -12m + 4n + 12 = 0

To solve these, we utilized substitution and elimination methods to find the values of \(m\) and \(n\).

Solving simultaneous equations involves algebraic manipulation to isolate and find the unknowns step-by-step.

polynomial functions

Polynomial functions are algebraic expressions that involve sums and products of variables and coefficients, raised to whole number exponents. These functions play a crucial role in various fields of mathematics and science.

The general form of a polynomial function is: \(a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\), where each \(a_i\) is a coefficient.

In our exercise, two polynomial functions were given:

• 2x^3 + mx^2 + nx - 3


• x^3 - 3mx^2 + 2nx + 4

The problem required us to determine the values of \(m\) and \(n\) so that both polynomials are divisible by \(x-2\). This involved applying the Remainder Theorem and solving the resulting simultaneous equations to make sure both polynomials evaluated to zero at \(x=2\).

substitution method

The substitution method is a strategy for solving systems of equations. It involves solving one equation for one variable and then substituting that expression into another equation. This reduces the number of variables and equations, making it simpler to solve for unknowns.

In our exercise, after establishing the simultaneous equations: \(4m + 2n + 13 = 0\) and \(-12m + 4n + 12 = 0\), we used substitution to isolate and solve for \(m\) and \(n\).

We first multiplied the first equation to match the coefficients of \(n\) in both equations. We then subtracted to eliminate \(n\), finding \(m\). Next, we substituted the value of \(m\) back into the original equation to find \(n\).

• Solve one equation for a chosen variable.


• Substitute the expression for that variable into the other equation.


• Solve the resulting single-variable equation.


• Back-substitute the found value into one of the original equations.
• Check the solutions against both original equations.

This method effectively reduces the complexity by focusing on one variable at a time.

divisibility conditions

Divisibility conditions in polynomials determine when a polynomial can be exactly divided by another polynomial, leaving no remainder. These conditions are vital in factorization, simplifying expressions, and solving polynomial equations.

In our exercise, we needed to find \(m\) and \(n\) such that both given polynomials are divisible by \(x-2\). This means the condition for divisibility by \(x-2\) is that substituting \(x=2\) into both polynomials should result in zero.

By applying the Remainder Theorem, which states a polynomial \(f(x)\) is divisible by \(x - a\) if and only if \(f(a) = 0\), we ensured both polynomials gave zero results upon substitution.

Hence, accurate application of divisibility conditions and knowledge of the Remainder Theorem guided us in solving for \(m\) and \(n\).

• Substitute the divisor value into the polynomial.


• Set the polynomial equation to zero.


• Solve for the unknowns to meet the no-remainder condition.

This process helps with polynomial factorization and simplification in algebraic equations.