## Step by step solution :

## Step 1 :

#### Equation at the end of step 1 :

` ((((r`^{4})+(r^{3}))-7r^{2})-r)+6 = 0

## Step 2 :

### Polynomial Roots Calculator :

2.1 Find roots (zeroes) of : F(r) = r^{4}+r^{3}-7r^{2}-r+6

Polynomial Roots Calculator is a set of methods aimed at finding values ofrfor which F(r)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers r which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational numberP/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 1 and the Trailing Constant is 6. The factor(s) are:

of the Leading Coefficient : 1

of the Trailing Constant : 1 ,2 ,3 ,6 Let us test ....

P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|

-1 | 1 | -1.00 | 0.00 | r+1 | |||||

-2 | 1 | -2.00 | -12.00 | ||||||

-3 | 1 | -3.00 | 0.00 | r+3 | |||||

-6 | 1 | -6.00 | 840.00 | ||||||

1 | 1 | 1.00 | 0.00 | r-1 | |||||

2 | 1 | 2.00 | 0.00 | r-2 | |||||

3 | 1 | 3.00 | 48.00 | ||||||

6 | 1 | 6.00 | 1260.00 |

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that

r^{4}+r^{3}-7r^{2}-r+6

can be divided by 4 different polynomials,including by r-2

### Polynomial Long Division :

2.2 Polynomial Long Division

Dividing : r^{4}+r^{3}-7r^{2}-r+6

("Dividend")

By:r-2("Divisor")

dividend | r^{4} | + | r^{3} | - | 7r^{2} | - | r | + | 6 | ||

-divisor | * r^{3} | r^{4} | - | 2r^{3} | |||||||

remainder | 3r^{3} | - | 7r^{2} | - | r | + | 6 | ||||

-divisor | * 3r^{2} | 3r^{3} | - | 6r^{2} | |||||||

remainder | - | r^{2} | - | r | + | 6 | |||||

-divisor | * -r^{1} | - | r^{2} | + | 2r | ||||||

remainder | - | 3r | + | 6 | |||||||

-divisor | * -3r^{0} | - | 3r | + | 6 | ||||||

remainder | 0 |

Quotient : r^{3}+3r^{2}-r-3 Remainder: 0

### Polynomial Roots Calculator :

2.3 Find roots (zeroes) of : F(r) = r^{3}+3r^{2}-r-3

See theory in step 2.1

In this case, the Leading Coefficient is 1 and the Trailing Constant is -3. The factor(s) are:

of the Leading Coefficient : 1

of the Trailing Constant : 1 ,3 Let us test ....

P | Q | P/Q | F(P/Q) | Divisor | |||||
---|---|---|---|---|---|---|---|---|---|

-1 | 1 | -1.00 | 0.00 | r+1 | |||||

-3 | 1 | -3.00 | 0.00 | r+3 | |||||

1 | 1 | 1.00 | 0.00 | r-1 | |||||

3 | 1 | 3.00 | 48.00 |

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that

r^{3}+3r^{2}-r-3

can be divided by 3 different polynomials,including by r-1

### Polynomial Long Division :

2.4 Polynomial Long Division

Dividing : r^{3}+3r^{2}-r-3

("Dividend")

By:r-1("Divisor")

dividend | r^{3} | + | 3r^{2} | - | r | - | 3 | ||

-divisor | * r^{2} | r^{3} | - | r^{2} | |||||

remainder | 4r^{2} | - | r | - | 3 | ||||

-divisor | * 4r^{1} | 4r^{2} | - | 4r | |||||

remainder | 3r | - | 3 | ||||||

-divisor | * 3r^{0} | 3r | - | 3 | |||||

remainder | 0 |

Quotient : r^{2}+4r+3 Remainder: 0

#### Trying to factor by splitting the middle term

2.5Factoring r^{2}+4r+3

The first term is, r^{2} its coefficient is 1.

The middle term is, +4r its coefficient is 4.

The last term, "the constant", is +3

Step-1 : Multiply the coefficient of the first term by the constant 1•3=3

Step-2 : Find two factors of 3 whose sum equals the coefficient of the middle term, which is 4.

-3 | + | -1 | = | -4 | ||

-1 | + | -3 | = | -4 | ||

1 | + | 3 | = | 4 | That's it |

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step2above, 1 and 3

r^{2} + 1r+3r + 3

Step-4 : Add up the first 2 terms, pulling out like factors:

r•(r+1)

Add up the last 2 terms, pulling out common factors:

3•(r+1)

Step-5:Add up the four terms of step4:

(r+3)•(r+1)

Which is the desired factorization

#### Equation at the end of step 2 :

` (r + 3) • (r + 1) • (r - 1) • (r - 2) = 0 `

## Step 3 :

#### Theory - Roots of a product :

3.1 A product of several terms equals zero.When a product of two or more terms equals zero, then at least one of the terms must be zero.We shall now solve each term = 0 separatelyIn other words, we are going to solve as many equations as there are terms in the productAny solution of term = 0 solves product = 0 as well.

#### Solving a Single Variable Equation:

3.2Solve:r+3 = 0Subtract 3 from both sides of the equation:

r = -3

#### Solving a Single Variable Equation:

3.3Solve:r+1 = 0Subtract 1 from both sides of the equation:

r = -1

#### Solving a Single Variable Equation:

3.4Solve:r-1 = 0Add 1 to both sides of the equation:

r = 1

#### Solving a Single Variable Equation:

3.5Solve:r-2 = 0Add 2 to both sides of the equation:

r = 2

### Supplement : Solving Quadratic Equation Directly

`Solving r`^{2}+4r+3 = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

#### Parabola, Finding the Vertex:

4.1Find the Vertex ofy = r^{2}+4r+3Parabolas have a highest or a lowest point called the Vertex.Our parabola opens up and accordingly has a lowest point (AKA absolute minimum).We know this even before plotting "y" because the coefficient of the first term,1, is positive (greater than zero).Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x-intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.For any parabola,Ar^{2}+Br+C,the r-coordinate of the vertex is given by -B/(2A). In our case the r coordinate is -2.0000Plugging into the parabola formula -2.0000 for r we can calculate the y-coordinate:

y = 1.0 * -2.00 * -2.00 + 4.0 * -2.00 + 3.0

or y = -1.000

#### Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = r^{2}+4r+3

Axis of Symmetry (dashed) {r}={-2.00}

Vertex at {r,y} = {-2.00,-1.00}

r-Intercepts (Roots) :

Root 1 at {r,y} = {-3.00, 0.00}

Root 2 at {r,y} = {-1.00, 0.00}

#### Solve Quadratic Equation by Completing The Square

4.2Solvingr^{2}+4r+3 = 0 by Completing The Square.Subtract 3 from both side of the equation :

r^{2}+4r = -3

Now the clever bit: Take the coefficient of r, which is 4, divide by two, giving 2, and finally square it giving 4

Add 4 to both sides of the equation :

On the right hand side we have:

-3+4or, (-3/1)+(4/1)

The common denominator of the two fractions is 1Adding (-3/1)+(4/1) gives 1/1

So adding to both sides we finally get:

r^{2}+4r+4 = 1

Adding 4 has completed the left hand side into a perfect square :

r^{2}+4r+4=

(r+2)•(r+2)=

(r+2)^{2}

Things which are equal to the same thing are also equal to one another. Since

r^{2}+4r+4 = 1 and

r^{2}+4r+4 = (r+2)^{2}

then, according to the law of transitivity,

(r+2)^{2} = 1

We'll refer to this Equation as Eq. #4.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(r+2)^{2} is

(r+2)^{2/2}=

(r+2)^{1}=

r+2

Now, applying the Square Root Principle to Eq.#4.2.1 we get:

r+2= √ 1

Subtract 2 from both sides to obtain:

r = -2 + √ 1

Since a square root has two values, one positive and the other negative

r^{2} + 4r + 3 = 0

has two solutions:

r = -2 + √ 1

or

r = -2 - √ 1

### Solve Quadratic Equation using the Quadratic Formula

4.3Solvingr^{2}+4r+3 = 0 by the Quadratic Formula.According to the Quadratic Formula,r, the solution forAr^{2}+Br+C= 0 , where A, B and C are numbers, often called coefficients, is given by :

-B± √B^{2}-4AC

r = ————————

2A In our case,A= 1

B= 4

C= 3 Accordingly,B^{2}-4AC=

16 - 12 =

4Applying the quadratic formula :

-4 ± √ 4

r=—————

2Can √ 4 be simplified ?

Yes!The prime factorization of 4is

2•2

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 4 =√2•2 =

±2 •√ 1 =

±2

So now we are looking at:

r=(-4±2)/2

Two real solutions:

r =(-4+√4)/2=-2+= -1.000

or:

r =(-4-√4)/2=-2-= -3.000

## Four solutions were found :

- r = 2
- r = 1
- r = -1
- r = -3