Question

If the thrice repeated roots of equation \(x^{4}+a x^{3}+b x^{2}+c x-1=0\) is 1 , then \(a+b+2 c\) is equal to (1) 0 (2) 1 (3) \(-1\) (4) 2

Short answer

-6

Step-by-step solution

Understand the Given Information

The given polynomial is: \(x^4 + a x^3 + b x^2 + c x - 1 = 0\)We are also given that the root 1 is repeated thrice.

Represent the Polynomial with Given Roots

Since 1 is a thrice repeated root, and this is a fourth degree polynomial, we can represent the polynomial as:\((x-1)^3(x-r) = x^4 + a x^3 + b x^2 + c x - 1\)where \(1\) is repeated thrice and \(r\) is an unknown root.

Expand the Polynomial

Expand \((x-1)^3(x-r)\):\((x^3 - 3x^2 + 3x - 1)(x - r)\)Expanding this:\[\begin{aligned} (x^3 - 3x^2 + 3x - 1)(x - r) &= x^4 - rx^3 - 3x^3 + 3rx^2 + 3x^2 - 3xr - x + r \ &= x^4 - (r + 3)x^3 + (3r + 3)x^2 - (3r + 1)x + r\end{aligned}\]

Match the Coefficients

Comparing the expanded polynomial to the given polynomial:\(x^4 - (r + 3)x^3 + (3r + 3)x^2 - (3r + 1)x + r = x^4 + a x^3 + b x^2 + c x - 1\)Matching coefficients, we get:\[a = -(r + 3)\]\[b = 3r + 3\]\[c = -(3r + 1)\]Also, comparing constant terms we get:\[r = 1\]

Substitute and Solve for a, b, and c

Using \(r = 1\):\[a = -(1 + 3) = -4\]\[b = 3(1) + 3 = 6\]\[c = -(3(1) + 1) = -4\]

Calculate \(a + b + 2c\)

Now, calculate \(a + b + 2c\):\[a + b + 2c = -4 + 6 + 2(-4) = -4 + 6 - 8 = -6\]

Key concepts

Polynomial Equations

Polynomial equations are mathematical expressions involving one or more variables and coefficients. They are expressed in the form poly(x) = 0, where poly(x) represents a polynomial in the variable x. Polynomials can have varying degrees, which indicate the highest power of the variable present. For example, the polynomial in our exercise, x^4 + ax^3 + bx^2 + cx - 1 = 0, is a fourth-degree polynomial. This is because the highest power of x is 4. Polynomial equations are solved by finding the values of the variable that make the equation true. These values are called the roots or solutions of the polynomial. For example, if x = 1 makes the polynomial x^2 - 3x + 2 = 0 true, then x = 1 is a root of the polynomial.
To solve polynomial equations, we often use techniques such as:

Factoring the polynomial

Using the Rational Root Theorem

Applying synthetic division

In our exercise, we use the fact that 1 is a thrice repeated root and express the polynomial accordingly to find the unknowns.

Root Multiplicity

Root multiplicity in polynomial equations refers to the number of times a particular root is repeated. If a root appears more than once, we say it has a higher multiplicity. In simpler terms, it means the root 'counts' as multiple roots. For instance, in the polynomial (x - 1)^3(x - r), the root x = 1 has a multiplicity of three, because it is repeated three times. Multiplicity provides important information about the behavior of the polynomial graph and its derivative. If a root has an odd multiplicity, the graph of the polynomial crosses the x-axis at that root. If the multiplicity is even, the graph touches the x-axis and turns back without crossing it.
In our exercise, we know that 1 is a thrice repeated root of the polynomial x^4 + ax^3 + bx^2 + cx - 1 = 0. This information allows us to set up the polynomial as (x - 1)^3(x - r) and then expand and match coefficients to solve for the unknowns.

Coefficient Comparison

Coefficient comparison is a method used in algebra to identify unknown coefficients in polynomials by matching terms of equal degree. This involves expanding both sides of an equation and then equating the coefficients of corresponding powers of variables. It's a powerful technique for solving polynomial equations and can be especially useful when dealing with roots of higher multiplicities. For example, in our polynomial, we start with an expression (x - 1)^3(x - r) and expand it:
(x^3 - 3x^2 + 3x - 1)(x - r) = x^4 - (r + 3)x^3 + (3r + 3)x^2 - (3r + 1)x + r
We then compare this expanded form to the given polynomial x^4 + ax^3 + bx^2 + cx - 1. By aligning the coefficients for each power of x, we get:
a = -(r + 3)
b = 3r + 3
c = -(3r + 1)
and solve for r. Once r is found to be 1, we substitute back to find the values of a, b, and c. This method ensures that we accurately determine the coefficients of the polynomial.