x = -1
x =(1-√5)/-2= 0.618
x =(1+√5)/-2=-1.618
Step 1 :
Equation at the end of step 1 :
0 - (((x3) + 2x2) - 1) = 0
Step 2 :
Step 3 :
Pulling out like terms :
3.1 Pull out like factors :
-x3 - 2x2 + 1 = -1 • (x3 + 2x2 - 1)
3.2 Find roots (zeroes) of : F(x) = x3 + 2x2 - 1
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 integers
The 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 Coefficient
In this case, the Leading Coefficient is 1 and the Trailing Constant is -1.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1
Let us test
P Q P/Q F(P/Q) Divisor
-1 1 -1.00 0.00 x + 1
1 1 1.00 2.00
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 - 1
can be divided with x + 1
Polynomial Long Division :
3.3 Polynomial Long Division
Dividing : x3 + 2x2 - 1
("Dividend")
By : x + 1 ("Divisor")
dividend x3 + 2x2 - 1
- divisor * x2 x3 + x2
remainder x2 - 1
- divisor * x1 x2 + x
remainder - x - 1
- divisor * -x0 - x - 1
remainder 0
Quotient : x2+x-1 Remainder: 0
Trying to factor by splitting the middle term
3.4 Factoring 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 -1
Step-1 : Multiply the coefficient of the first term by the constant 1 • -1 = -1
Step-2 : Find two factors of -1 whose sum equals the coefficient of the middle term, which is 1 .
-1 + 1 = 0
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 3 :
(-x2 - x + 1) • (x + 1) = 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 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Parabola, Finding the Vertex :
4.2 Find the Vertex of y = -x2-x+1
For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is -0.5000
Plugging 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 = 1.250
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, 1.25}
x -Intercepts (Roots) :
Root 1 at {x,y} = { 0.62, 0.00}
Root 2 at {x,y} = {-1.62, 0.00}
Solve Quadratic Equation by Completing The Square
4.3 Solving -x2-x+1 = 0 by Completing The Square .
Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term:
x2+x-1 = 0 Add 1 to both side of the equation :
x2+x = 1
Now the clever bit: Take the coefficient of x , which is 1 , divide by two, giving 1/2 , and finally square it giving 1/4
Add 1/4 to both sides of the equation :
On the right hand side we have :
1 + 1/4 or, (1/1)+(1/4)
The common denominator of the two fractions is 4 Adding (4/4)+(1/4) gives 5/4
So adding to both sides we finally get :
x2+x+(1/4) = 5/4
Adding 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. Since
x2+x+(1/4) = 5/4 and
x2+x+(1/4) = (x+(1/2))2
then, according to the law of transitivity,
(x+(1/2))2 = 5/4
We'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+(1/2))2 is
(x+(1/2))2/2 =
(x+(1/2))1 =
x+(1/2)
Now, applying the Square Root Principle to Eq. #4.3.1 we get:
x+(1/2) = √ 5/4
Subtract 1/2 from both sides to obtain:
x = -1/2 + √ 5/4
Since a square root has two values, one positive and the other negative
x2 + x - 1 = 0
has two solutions:
x = -1/2 + √ 5/4
or
x = -1/2 - √ 5/4
Note that √ 5/4 can be written as
√ 5 / √ 4 which is √ 5 / 2
Solve Quadratic Equation using the Quadratic Formula
4.4 Solving -x2-x+1 = 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 = -1
C = 1
Accordingly, B2 - 4AC =
1 - (-4) =
5
Applying the quadratic formula :
1 ± √ 5
x = ————
-2
√ 5 , rounded to 4 decimal digits, is 2.2361
So now we are looking at:
x = ( 1 ± 2.236 ) / -2
Two real solutions:
x =(1+√5)/-2=-1.618
or:
x =(1-√5)/-2= 0.618
Solving a Single Variable Equation :
4.5 Solve : x+1 = 0
Subtract 1 from both sides of the equation :
x = -1
Hope this helps.
