Solve Polynomialrootcalculator 2x^3+5x^2-8x-15=0 ... - Tiger Algebra

Step by step solution :

Step 1 :

Equation at the end of step 1 :

(((2 • (x3)) + 5x2) - 8x) - 15 = 0

Step 2 :

Equation at the end of step 2 :

((2x3 + 5x2) - 8x) - 15 = 0

Step 3 :

Checking for a perfect cube :

3.1 2x3+5x2-8x-15 is not a perfect cube

Trying to factor by pulling out :

3.2 Factoring: 2x3+5x2-8x-15 Thoughtfully split the expression at hand into groups, each group having two terms :Group 1: -8x-15 Group 2: 2x3+5x2 Pull out from each group separately :Group 1: (8x+15) • (-1)Group 2: (2x+5) • (x2)Bad news !! Factoring by pulling out fails : The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

3.3 Find roots (zeroes) of : F(x) = 2x3+5x2-8x-15Polynomial Roots Calculator is a set of methods aimed at finding values of x for 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 number P/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 2 and the Trailing Constant is -15. The factor(s) are: of the Leading Coefficient : 1,2 of the Trailing Constant : 1 ,3 ,5 ,15 Let us test ....

PQP/QF(P/Q)Divisor
-1 1 -1.00 -4.00
-1 2 -0.50 -10.00
-3 1 -3.00 0.00 x+3
-3 2 -1.50 1.50
-5 1 -5.00 -100.00
-5 2 -2.50 5.00
-15 1 -15.00 -5520.00
-15 2 -7.50 -517.50
1 1 1.00 -16.00
1 2 0.50 -17.50
3 1 3.00 60.00
3 2 1.50 -9.00
5 1 5.00 320.00
5 2 2.50 27.50
15 1 15.00 7740.00
15 2 7.50 1050.00

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 2x3+5x2-8x-15 can be divided with x+3

Polynomial Long Division :

3.4 Polynomial Long Division Dividing : 2x3+5x2-8x-15 ("Dividend") By : x+3 ("Divisor")

dividend 2x3 + 5x2 - 8x - 15
- divisor * 2x2 2x3 + 6x2
remainder- x2 - 8x - 15
- divisor * -x1 - x2 - 3x
remainder- 5x - 15
- divisor * -5x0 - 5x - 15
remainder0

Quotient : 2x2-x-5 Remainder: 0

Trying to factor by splitting the middle term

3.5 Factoring 2x2-x-5 The first term is, 2x2 its coefficient is 2 .The middle term is, -x its coefficient is -1 .The last term, "the constant", is -5 Step-1 : Multiply the coefficient of the first term by the constant 2-5 = -10 Step-2 : Find two factors of -10 whose sum equals the coefficient of the middle term, which is -1 .

-10 + 1 = -9
-5 + 2 = -3
-2 + 5 = 3
-1 + 10 = 9

Observation : No two such factors can be found !! Conclusion : Trinomial can not be factored

Equation at the end of step 3 :

(2x2 - x - 5) • (x + 3) = 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.

Parabola, Finding the Vertex :

4.2 Find the Vertex of y = 2x2-x-5Parabolas 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, 2 , 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,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 0.2500 Plugging into the parabola formula 0.2500 for x we can calculate the y -coordinate : y = 2.0 * 0.25 * 0.25 - 1.0 * 0.25 - 5.0 or y = -5.125

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = 2x2-x-5 Axis of Symmetry (dashed) {x}={ 0.25} Vertex at {x,y} = { 0.25,-5.12} x -Intercepts (Roots) : Root 1 at {x,y} = {-1.35, 0.00} Root 2 at {x,y} = { 1.85, 0.00}

Solve Quadratic Equation by Completing The Square

4.3 Solving 2x2-x-5 = 0 by Completing The Square . Divide both sides of the equation by 2 to have 1 as the coefficient of the first term : x2-(1/2)x-(5/2) = 0Add 5/2 to both side of the equation : x2-(1/2)x = 5/2Now the clever bit: Take the coefficient of x , which is 1/2 , divide by two, giving 1/4 , and finally square it giving 1/16 Add 1/16 to both sides of the equation : On the right hand side we have : 5/2 + 1/16 The common denominator of the two fractions is 16 Adding (40/16)+(1/16) gives 41/16 So adding to both sides we finally get : x2-(1/2)x+(1/16) = 41/16Adding 1/16 has completed the left hand side into a perfect square : x2-(1/2)x+(1/16) = (x-(1/4)) (x-(1/4)) = (x-(1/4))2 Things which are equal to the same thing are also equal to one another. Since x2-(1/2)x+(1/16) = 41/16 and x2-(1/2)x+(1/16) = (x-(1/4))2 then, according to the law of transitivity, (x-(1/4))2 = 41/16We'll refer to this Equation as Eq. #4.3.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of (x-(1/4))2 is (x-(1/4))2/2 = (x-(1/4))1 = x-(1/4)Now, applying the Square Root Principle to Eq. #4.3.1 we get: x-(1/4) = 41/16 Add 1/4 to both sides to obtain: x = 1/4 + √ 41/16 Since a square root has two values, one positive and the other negative x2 - (1/2)x - (5/2) = 0 has two solutions: x = 1/4 + √ 41/16 or x = 1/4 - √ 41/16 Note that 41/16 can be written as 41 / √ 16 which is 41 / 4

Solve Quadratic Equation using the Quadratic Formula

4.4 Solving 2x2-x-5 = 0 by the Quadratic Formula . According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by : - B ± √ B2-4AC x = ———————— 2A In our case, A = 2 B = -1 C = -5 Accordingly, B2 - 4AC = 1 - (-40) = 41Applying the quadratic formula : 1 ± √ 41 x = ————— 4 41 , rounded to 4 decimal digits, is 6.4031 So now we are looking at: x = ( 1 ± 6.403 ) / 4Two real solutions: x =(1+√41)/4= 1.851 or: x =(1-√41)/4=-1.351

Solving a Single Variable Equation :

4.5 Solve : x+3 = 0Subtract 3 from both sides of the equation : x = -3

Three solutions were found :

  1. x = -3
  2. x =(1-√41)/4=-1.351
  3. x =(1+√41)/4= 1.851

Từ khóa » F(x)=x^2-8x+15 G(x)=x-3