Finding Roots - Polynomial

Category: Algebra

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Find the roots for the given polynomial:

x⁵ - 3x⁴ - x³ + 11x² - 12x + 4 = 0

Solution:

To solve for the roots of a given polynomial, we have to see the last term which is 4. Let's assign the factors of 4 which are 1, -1, 2, -2, 4, and -4.

First, let's check the number of positive and negative roots of a given polynomial using Descartes' Rule of Sign. Without changing the sign of x, the number of positive roots are 4 as shown below:

Next change x into -x and substitute to the given equation:

(-x)⁵ - 3(-x)⁴ - (-x)³ + 11(-x)² - 12(-x) + 4 = 0

-x⁵ - 3x⁴ + x³ + 11x² + 12x + 4 = 0

Since the highest degree of a given polynomial is 5, the total number of roots must be 5. There are 4 positive roots and 1 negative root of a given polynomial by Descartes' Rule of Sign. Let's see if the number of roots are correct by getting the factors using Synthetic Division, we have

x⁵ - 3x4 - x³ + 11x² - 12x + 4 = 0

If you will perform the Synthetic Division method, you have to arrange the polynomial in descending order first and then consider only their coefficient. In case that any of their middle term is missing, then you have to consider 0 as the coefficient of the missing term.

Continue to do the Synthetic Division as much as you can until you have one term left on the left side. You have to do this in trial and error using the factors of the last term so that the remainder is zero at the right side.

Therefore, the roots are 1, 1, 1, 2, and -2. 

Derivative - Increment Method

Category: Differential Calculus

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Find the derivative of the given equation using the Increment Method:

Solution:

This is the original method in finding the derivative of any equations using the Increment Method. If you know already the List of Formulas to find a derivative of a certain equation, then you can use it. If you don't remember the formulas, then you have to use the Increment Method and it is a long process to do it. Well, for the Increment Method, you have to substitute x with x + Δx and substitute y with y + Δy, 

Subtract the given equation from the above equation, we have,

Get the Least Common Denominator (LCD) of the two terms of the right,

Simplify the above equation,

Divide both sides of the equation by Δx,

Therefore,

The above equation can be written as

Proving - Congruent Triangles

Category: Plane Geometry

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Given the figure below:

Photo by Math Principles in Everyday Life

1. Given AF ≅ AD, and FE ≅ ED. Prove that ΔAFE ≅ ΔAED.

2. Given ABDE is a square and ΔACD is an isosceles triangle.  Prove that ΔAED ≅ ΔBDC.

3. Given ABDE is a square. Prove that ΔABD ≅ ΔADE.

4. Given AB ≅ DE, AD bisects ∠BAE and ∠BDE. Prove that ΔBAD ≅ ΔADE.

5. Given AF ≅ DC and DC ≅ ED. Prove that ΔAFD ≅ ΔACD.

Solution:

Consider Case 1:

Photo by Math Principles in Everyday Life

Proof:

  

1. Statement: AF ≅ AD and FE ≅ ED.

    Reason: Given items.

2. Statement: AE ≅ AE. 

    Reason: Reflexive property of congruence.

Therefore, ΔAFE ≅ ΔAED.

Reason: SSS (Side-Side-Side) Postulate

Consider Case 2:

Photo by Math Principles in Everyday Life

Proof:

1. Statement: ABDE is a square and ΔACD is an isosceles triangle.

    Reason: Given items.

2. Statement: AE ≅ BD ≅ AB ≅ ED

    Reason: All sides of a square are congruent.

3. Statement: AD ≅ AD

    Reason: Reflexive property of congruence.

4. Statement: ∠EDA ≅ ∠BAD

    Reason: The alternating interior angles of a two parallel lines that passes a transversal line are congruent. The two opposite sides of a square are parallel.

5. Statement: ∠DAB ≅ ∠BCD

    Reason: The base angles of an isosceles triangle are congruent.

6. Statement: AD ≅ DC

    Reason: The two sides of an isosceles triangle are congruent.

7. Statement: AB ≅ BC

    Reason: The base altitude (BD) of an isosceles triangle bisects the line segment (AC) of a base.

Therefore, ΔAED ≅ ΔBDC.

Reason: SAS (Side-Angle-Side) Postulate

Consider Case 3:

Photo by Math Principles in Everyday Life

Proof:

1. Statement: ABDE is a square.

    Reason: Given item.

2. Statement: AE ≅ BD ≅ AB ≅ ED

    Reason: All sides of a square are congruent.

3. Statement: AD ≅ AD

    Reason: Reflexive property of congruence.

Therefore, ΔABD ≅ ΔAED.

Reason: SSS (Side-Side-Side) Postulate

Consider Case 4:

Photo by Math Principles in Everyday Life

Proof:

1. Statement: AB ≅ ED, ∠EAD ≅ ∠BAD, and ∠BDA ≅ ∠ADE.

    Reason: Given items.

2. Statement: AD ≅ AD

    Reason: Reflexive property of congruence.

Therefore, ΔBAD ≅ ΔADE.

Reason, ASA (Angle-Side-Angle) Postulate.

Consider Case 5:

Photo by Math Principles in Everyday Life

Proof:

1. Statement: AF ≅ DC and DC ≅ ED.

    Reason: Given items.

2. Statement: AB ≅ ED and AE ≅ BD.

    Reason: The opposite sides of a rectangle are congruent.

3. Statement: AD ≅ AD

    Reason: Reflexive property of congruence.

4. Statement: AE ┴ ED and BD ┴ AB.

    Reason: The sides of a rectangle ABDE are perpendicular to each other.

5. Statement: ∠FEA ≅ ∠AED, and ∠ABD ≅ ∠DBC.

    Reason: The sum of the supplementary angles is 180. If ∠AED and ∠ABD are 90°, then ∠FEA and ∠DBC msut be 90°.

6. Statement: ΔFEA and ΔDBC are right triangles.

    Reason: One of the angles of each triangles is 90°.

7. Statement: FE ≅ BC

    Reason: Since ΔFEA and ΔDBC are right triangles, we can use Pythagorean Theorem (c² = a² + b²) to solve the other side of a right triangle. If AF ≅ CD and AE ≅ BD, then FE ≅ BC. 

8. Statement: FD ≅ AC

    Reason: Since FE ≅ BC and ED ≅ AB, then FE + ED ≅ AB + BC. 

Therefore, ΔAFD ≅ ΔACD.

Reason: SSS (Side-Side-Side) Postulate