Solve Polynomiallongdivision X3+x2-21x-45=0 Tiger Algebra Solver

Reformatting the input :

Changes made to your input should not affect the solution: (1): "x2" was replaced by "x^2". 1 more similar replacement(s).

Step by step solution :

Step 1 :

Checking for a perfect cube :

1.1 x3+x2-21x-45 is not a perfect cube

Trying to factor by pulling out :

1.2 Factoring: x3+x2-21x-45 Thoughtfully split the expression at hand into groups, each group having two terms :Group 1: -21x-45 Group 2: x3+x2 Pull out from each group separately :Group 1: (7x+15) • (-3)Group 2: (x+1) • (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 :

1.3 Find roots (zeroes) of : F(x) = x3+x2-21x-45Polynomial 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 1 and the Trailing Constant is -45. The factor(s) are: of the Leading Coefficient : 1 of the Trailing Constant : 1 ,3 ,5 ,9 ,15 ,45 Let us test ....

P	Q	P/Q	F(P/Q)	Divisor
-1	1	-1.00	-24.00
-3	1	-3.00	0.00	x+3
-5	1	-5.00	-40.00
-9	1	-9.00	-504.00
-15	1	-15.00	-2880.00
-45	1	-45.00	-88200.00
1	1	1.00	-64.00
3	1	3.00	-72.00
5	1	5.00	0.00	x-5
9	1	9.00	576.00
15	1	15.00	3240.00
45	1	45.00	92160.00

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+x2-21x-45 can be divided by 2 different polynomials,including by x-5

Polynomial Long Division :

1.4 Polynomial Long Division Dividing : x3+x2-21x-45 ("Dividend") By : x-5 ("Divisor")

dividend	x3	+	x2	-	21x	-	45
- divisor	* x2	x3	-	5x2
remainder	6x2	-	21x	-	45
- divisor	* 6x1	6x2	-	30x
remainder	9x	-	45
- divisor	* 9x0	9x	-	45
remainder	0

Quotient : x2+6x+9 Remainder: 0

Trying to factor by splitting the middle term

1.5 Factoring x2+6x+9 The first term is, x2 its coefficient is 1 .The middle term is, +6x its coefficient is 6 .The last term, "the constant", is +9 Step-1 : Multiply the coefficient of the first term by the constant 1 • 9 = 9 Step-2 : Find two factors of 9 whose sum equals the coefficient of the middle term, which is 6 .

-9	+	-1	=	-10
-3	+	-3	=	-6
-1	+	-9	=	-10
1	+	9	=	10
3	+	3	=	6	That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 3 and 3 x2 + 3x + 3x + 9Step-4 : Add up the first 2 terms, pulling out like factors : x • (x+3) Add up the last 2 terms, pulling out common factors : 3 • (x+3) Step-5 : Add up the four terms of step 4 : (x+3) • (x+3) Which is the desired factorization

Multiplying Exponential Expressions :

1.6 Multiply (x+3) by (x+3) The rule says : To multiply exponential expressions which have the same base, add up their exponents.In our case, the common base is (x+3) and the exponents are : 1 , as (x+3) is the same number as (x+3)1 and 1 , as (x+3) is the same number as (x+3)1 The product is therefore, (x+3)(1+1) = (x+3)2

Equation at the end of step 1 :

(x + 3)2 • (x - 5) = 0

Step 2 :

Theory - Roots of a product :

2.1 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 :

2.2 Solve : (x+3)2 = 0 (x+3) 2 represents, in effect, a product of 2 terms which is equal to zero For the product to be zero, at least one of these terms must be zero. Since all these terms are equal to each other, it actually means : x+3 = 0 Subtract 3 from both sides of the equation : x = -3

Solving a Single Variable Equation :

2.3 Solve : x-5 = 0Add 5 to both sides of the equation : x = 5

Supplement : Solving Quadratic Equation Directly

Solving x2+6x+9 = 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 :

3.1 Find the Vertex of y = x2+6x+9Parabolas 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,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is -3.0000 Plugging into the parabola formula -3.0000 for x we can calculate the y -coordinate : y = 1.0 * -3.00 * -3.00 + 6.0 * -3.00 + 9.0 or y = 0.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2+6x+9 Vertex at {x,y} = {-3.00, 0.00} x-Intercept (Root) : One Root at {x,y}={-3.00, 0.00} Note that the root coincides with the Vertex and the Axis of Symmetry coinsides with the line x = 0

Solve Quadratic Equation by Completing The Square

3.2 Solving x2+6x+9 = 0 by Completing The Square . Subtract 9 from both side of the equation : x2+6x = -9Now the clever bit: Take the coefficient of x , which is 6 , divide by two, giving 3 , and finally square it giving 9 Add 9 to both sides of the equation : On the right hand side we have : -9 + 9 or, (-9/1)+(9/1) The common denominator of the two fractions is 1 Adding (-9/1)+(9/1) gives 0/1 So adding to both sides we finally get : x2+6x+9 = 0Adding 9 has completed the left hand side into a perfect square : x2+6x+9 = (x+3) • (x+3) = (x+3)2 Things which are equal to the same thing are also equal to one another. Since x2+6x+9 = 0 and x2+6x+9 = (x+3)2 then, according to the law of transitivity, (x+3)2 = 0We'll refer to this Equation as Eq. #3.2.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of (x+3)2 is (x+3)2/2 = (x+3)1 = x+3Now, applying the Square Root Principle to Eq. #3.2.1 we get: x+3 = √ 0 Subtract 3 from both sides to obtain: x = -3 + √ 0 The square root of zero is zero This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.The solution is: x = -3

Solve Quadratic Equation using the Quadratic Formula

3.3 Solving x2+6x+9 = 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 = 1 B = 6 C = 9 Accordingly, B2 - 4AC = 36 - 36 = 0Applying the quadratic formula : -6 ± √ 0 x = ————— 2The square root of zero is zero This quadratic equation has one solution only. That's because adding zero is the same as subtracting zero.The solution is: x = -6 / 2 = -3

Two solutions were found :

1. x = 5
2. x = -3