### x^4+8x^3+7x^2-40x-60=0

This deals with finding the roots (zeroes) of polynomials.

You are watching: Find the rational roots of x^4+8x^3+7x^2-40x-60=0

## Step by Step Solution

## Step by step solution :

## Step 1 :

Equation at the end of step 1 : ((((x4)+(8•(x3)))+7x2)-40x)-60 = 0## Step 2 :

Equation at the end of step 2 : ((((x4) + 23x3) + 7x2) - 40x) - 60 = 0## Step 3 :

### Polynomial Roots Calculator :

3.1 Find roots (zeroes) of : F(x) = x4+8x3+7x2-40x-60Polynomial Roots Calculator is a set of methods aimed at finding values ofxfor which F(x)=0 Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers x which can be expressed as the quotient of two integersThe 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 CoefficientIn this case, the Leading Coefficient is 1 and the Trailing Constant is -60. The factor(s) are: of the Leading Coefficient : 1of the Trailing Constant : 1 ,2 ,3 ,4 ,5 ,6 ,10 ,12 ,15 ,20 , etc Let us test ....PQP/QF(P/Q)Divisor-1 | 1 | -1.00 | -20.00 | ||||||

-2 | 1 | -2.00 | 0.00 | x+2 | |||||

-3 | 1 | -3.00 | -12.00 | ||||||

-4 | 1 | -4.00 | -44.00 | ||||||

-5 | 1 | -5.00 | -60.00 | ||||||

-6 | 1 | -6.00 | 0.00 | x+6 | |||||

-10 | 1 | -10.00 | 3040.00 | ||||||

-12 | 1 | -12.00 | 8340.00 | ||||||

-15 | 1 | -15.00 | 25740.00 | ||||||

-20 | 1 | -20.00 | 99540.00 | ||||||

1 | 1 | 1.00 | -84.00 | ||||||

2 | 1 | 2.00 | -32.00 | ||||||

3 | 1 | 3.00 | 180.00 | ||||||

4 | 1 | 4.00 | 660.00 | ||||||

5 | 1 | 5.00 | 1540.00 | ||||||

6 | 1 | 6.00 | 2976.00 | ||||||

10 | 1 | 10.00 | 18240.00 | ||||||

12 | 1 | 12.00 | 35028.00 | ||||||

15 | 1 | 15.00 | 78540.00 | ||||||

20 | 1 | 20.00 | 225940.00 |

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 x4+8x3+7x2-40x-60can be divided by 2 different polynomials,including by x+6

### Polynomial Long Division :

3.2 Polynomial Long Division Dividing : x4+8x3+7x2-40x-60("Dividend") By:x+6("Divisor")

dividend | x4 | + | 8x3 | + | 7x2 | - | 40x | - | 60 | ||

-divisor | * x3 | x4 | + | 6x3 | |||||||

remainder | 2x3 | + | 7x2 | - | 40x | - | 60 | ||||

-divisor | * 2x2 | 2x3 | + | 12x2 | |||||||

remainder | - | 5x2 | - | 40x | - | 60 | |||||

-divisor | * -5x1 | - | 5x2 | - | 30x | ||||||

remainder | - | 10x | - | 60 | |||||||

-divisor | * -10x0 | - | 10x | - | 60 | ||||||

remainder | 0 |

Quotient : x3+2x2-5x-10 Remainder: 0

### Polynomial Roots Calculator :

3.3 Find roots (zeroes) of : F(x) = x3+2x2-5x-10See theory in step 3.1 In this case, the Leading Coefficient is 1 and the Trailing Constant is -10. The factor(s) are: of the Leading Coefficient : 1of the Trailing Constant : 1 ,2 ,5 ,10 Let us test ....

PQP/QF(P/Q)Divisor-1 | 1 | -1.00 | -4.00 | ||||||

-2 | 1 | -2.00 | 0.00 | x+2 | |||||

-5 | 1 | -5.00 | -60.00 | ||||||

-10 | 1 | -10.00 | -760.00 | ||||||

1 | 1 | 1.00 | -12.00 | ||||||

2 | 1 | 2.00 | -4.00 | ||||||

5 | 1 | 5.00 | 140.00 | ||||||

10 | 1 | 10.00 | 1140.00 |

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 x3+2x2-5x-10can be divided with x+2

### Polynomial Long Division :

3.4 Polynomial Long Division Dividing : x3+2x2-5x-10("Dividend") By:x+2("Divisor")

dividend | x3 | + | 2x2 | - | 5x | - | 10 | ||

-divisor | * x2 | x3 | + | 2x2 | |||||

remainder | - | 5x | - | 10 | |||||

-divisor | * 0x1 | ||||||||

remainder | - | 5x | - | 10 | |||||

-divisor | * -5x0 | - | 5x | - | 10 | ||||

remainder | 0 |

Quotient : x2-5 Remainder: 0

Trying to factor as a Difference of Squares:3.5 Factoring: x2-5 Theory : A difference of two perfect squares, A2-B2can be factored into (A+B)•(A-B)Proof:(A+B)•(A-B)= A2 - AB+BA-B2= A2 -AB+ AB - B2 = A2 - B2Note : AB = BA is the commutative property of multiplication. Note : -AB+ AB equals zero and is therefore eliminated from the expression.Check: 5 is not a square !! Ruling : Binomial can not be factored as the difference of two perfect squares.

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(x2 - 5) • (x + 2) • (x + 6) = 0

## Step 4 :

Theory - Roots of a product :4.1 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:4.2Solve:x2-5 = 0Add 5 to both sides of the equation:x2 = 5 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get: x = ± √ 5 The equation has two real solutions These solutions are x = ± √5 = ± 2.2361

Solving a Single Variable Equation:4.3Solve:x+2 = 0Subtract 2 from both sides of the equation:x = -2

Solving a Single Variable Equation:4.4Solve:x+6 = 0Subtract 6 from both sides of the equation:x = -6