## Step 1 :

Trying to factor as a Difference of Cubes:1.1 Factoring: x3-1 Theory : A difference of two perfect cubes, a3-b3 can be factored into(a-b)•(a2+ab+b2)Proof:(a-b)•(a2+ab+b2)=a3+a2b+ab2-ba2-b2a-b3 =a3+(a2b-ba2)+(ab2-b2a)-b3=a3+0+0-b3=a3-b3Check:1is the cube of 1Check: x3 is the cube of x1Factorization is :(x - 1)•(x2 + x + 1)

Trying to factor by splitting the middle term

1.2Factoring x2 + x + 1 The first term is, x2 its coefficient is 1.The middle term is, +x its coefficient is 1.The last term, "the constant", is +1Step-1 : Multiply the coefficient of the first term by the constant 1•1=1Step-2 : Find two factors of 1 whose sum equals the coefficient of the middle term, which is 1.

 -1 + -1 = -2 1 + 1 = 2

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

Equation at the end of step 1 :

(x - 1) • (x2 + x + 1) = 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:2.2Solve:x-1 = 0Add 1 to both sides of the equation:x = 1

Parabola, Finding the Vertex:2.3Find the Vertex ofy = x2+x+1Parabolas 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 -0.5000Plugging into the parabola formula -0.5000 for x we can calculate the y-coordinate:y = 1.0 * -0.50 * -0.50 + 1.0 * -0.50 + 1.0 or y = 0.750

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2+x+1 Axis of Symmetry (dashed) {x}={-0.50} Vertex at {x,y} = {-0.50, 0.75} Function has no real roots

Solve Quadratic Equation by Completing The Square

2.4Solvingx2+x+1 = 0 by Completing The Square.Subtract 1 from both side of the equation :x2+x = -1Now the clever bit: Take the coefficient of x, which is 1, divide by two, giving 1/2, and finally square it giving 1/4Add 1/4 to both sides of the equation :On the right hand side we have:-1+1/4or, (-1/1)+(1/4)The common denominator of the two fractions is 4Adding (-4/4)+(1/4) gives -3/4So adding to both sides we finally get:x2+x+(1/4) = -3/4Adding 1/4 has completed the left hand side into a perfect square :x2+x+(1/4)=(x+(1/2))•(x+(1/2))=(x+(1/2))2 Things which are equal to the same thing are also equal to one another. Sincex2+x+(1/4) = -3/4 andx2+x+(1/4) = (x+(1/2))2 then, according to the law of transitivity,(x+(1/2))2 = -3/4We"ll refer to this Equation as Eq. #2.4.1 The Square Root Principle says that When two things are equal, their square roots are equal.Note that the square root of(x+(1/2))2 is(x+(1/2))2/2=(x+(1/2))1=x+(1/2)Now, applying the Square Root Principle to Eq.#2.4.1 we get:x+(1/2)= √ -3/4 Subtract 1/2 from both sides to obtain:x = -1/2 + √ -3/4 In Math,iis called the imaginary unit. It satisfies i2=-1. Both i and -i are the square roots of -1Since a square root has two values, one positive and the other negativex2 + x + 1 = 0has two solutions:x = -1/2 + √ 3/4 • iorx = -1/2 - √ 3/4 • iNote that √ 3/4 can be written as√3 / √4which is √3 / 2

2.5Solvingx2+x+1 = 0 by the Quadratic Formula.According to the Quadratic Formula,x, the solution forAx2+Bx+C= 0 , where A, B and C are numbers, often called coefficients, is given by :-B± √B2-4ACx = ————————2A In our case,A= 1B= 1C= 1 Accordingly,B2-4AC=1 - 4 =-3Applying the quadratic formula : -1 ± √ -3 x=—————2In the set of real numbers, negative numbers do not have square roots.

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A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)Both i and -i are the square roots of minus 1Accordingly,√-3=√3•(-1)=√3•√-1=±√ 3 •i √ 3 , rounded to 4 decimal digits, is 1.7321So now we are looking at:x=(-1± 1.732 i )/2Two imaginary solutions :

x =(-1+√-3)/2=(-1+i√ 3 )/2= -0.5000+0.8660ior: x =(-1-√-3)/2=(-1-i√ 3 )/2= -0.5000-0.8660i

## Three solutions were found :

x =(-1-√-3)/2=(-1-i√ 3 )/2= -0.5000-0.8660ix =(-1+√-3)/2=(-1+i√ 3 )/2= -0.5000+0.8660ix = 1