Which of the following statements about the polynomial function f(x)=x^3+2x^2-1

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.