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

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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
-11 -1.00 -20.00
-21 -2.00 0.00x+2
-31 -3.00 -12.00
-41 -4.00 -44.00
-51 -5.00 -60.00
-61 -6.00 0.00x+6
-101-10.00 3040.00
-121-12.00 8340.00
-151-15.0025740.00
-201-20.0099540.00
11 1.00 -84.00
21 2.00 -32.00
31 3.00 180.00
41 4.00 660.00
51 5.00 1540.00
61 6.00 2976.00
101 10.0018240.00
121 12.0035028.00
151 15.0078540.00
201 20.00225940.00

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")

dividendx4+8x3+7x2-40x-60
-divisor* x3x4+6x3
remainder2x3+7x2-40x-60
-divisor* 2x22x3+12x2
remainder-5x2-40x-60
-divisor* -5x1-5x2-30x
remainder-10x-60
-divisor* -10x0-10x-60
remainder0

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
-11 -1.00 -4.00
-21 -2.00 0.00x+2
-51 -5.00 -60.00
-101-10.00 -760.00
11 1.00 -12.00
21 2.00 -4.00
51 5.00 140.00
101 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")

dividendx3+2x2-5x-10
-divisor* x2x3+2x2
remainder-5x-10
-divisor* 0x1
remainder-5x-10
-divisor* -5x0-5x-10
remainder0

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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Equation at the end of step 3 :

(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