Question

if x=-4 is a zero of the polynomial function f(x)=2x3+13x2+23x+12 which of the following is another zero of f(x)?

Answer 1

x4-7x3+13x2+23x-78=0

Four solutions were found :

x = 3
x = -2
x =(6-√-16)/2=3-2i= 3.0000-2.0000i
x =(6+√-16)/2=3+2i= 3.0000+2.0000i
Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2" was replaced by "x^2". 2 more similar replacement(s).

Step by step solution :

Step 1 :

Simplify x4 - 7x3
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Equation at the end of step 1 :

(((x4 - 7x3) + 13x2) + 23x) - 78 = 0
Step 2 :

Simplify x4-7x3 + 13x2
Equation at the end of step 2 :

((x4 - 7x3 + 13x2) + 23x) - 78 = 0
Step 3 :

Simplify x4-7x3+13x2 + 23x
Equation at the end of step 3 :

(x4 - 7x3 + 13x2 + 23x) - 78 = 0
Step 4 :

Simplify x4-7x3+13x2+23x - 78
Polynomial Roots Calculator :

4.1 Find roots (zeroes) of : F(x) = x4-7x3+13x2+23x-78
Polynomial 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 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 -78.

The factor(s) are:

of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,6 ,13 ,26 ,39 ,78

Let us test ....

P Q P/Q F(P/Q) Divisor
-1 1 -1.00 -80.00
-2 1 -2.00 0.00 x+2
-3 1 -3.00 240.00
-6 1 -6.00 3060.00
-13 1 -13.00 45760.00
-26 1 -26.00 588120.00
-39 1 -39.00 2747472.00
-78 1 -78.00 40414140.00
1 1 1.00 -48.00
2 1 2.00 -20.00
3 1 3.00 0.00 x-3
6 1 6.00 312.00
13 1 13.00 15600.00
26 1 26.00 343252.00
39 1 39.00 1918800.00
78 1 78.00 33774000.00

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
x4-7x3+13x2+23x-78
can be divided by 2 different polynomials,including by x-3

Polynomial Long Division :

4.2 Polynomial Long Division
Dividing : x4-7x3+13x2+23x-78
("Dividend")
By : x-3 ("Divisor")

dividend x4 - 7x3 + 13x2 + 23x - 78
- divisor * x3 x4 - 3x3
remainder - 4x3 + 13x2 + 23x - 78
- divisor * -4x2 - 4x3 + 12x2
remainder x2 + 23x - 78
- divisor * x1 x2 - 3x
remainder 26x - 78
- divisor * 26x0 26x - 78
remainder 0
Quotient : x3-4x2+x+26 Remainder: 0

Polynomial Roots Calculator :

4.3 Find roots (zeroes) of : F(x) = x3-4x2+x+26

See theory in step 4.1
In this case, the Leading Coefficient is 1 and the Trailing Constant is 26.

The factor(s) are:

of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,13 ,26

Let us test ....

P Q P/Q F(P/Q) Divisor
-1 1 -1.00 20.00
-2 1 -2.00 0.00 x+2
-13 1 -13.00 -2860.00
-26 1 -26.00 -20280.00
1 1 1.00 24.00
2 1 2.00 20.00
13 1 13.00 1560.00
26 1 26.00 14924.00

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
x3-4x2+x+26
can be divided with x+2

Polynomial Long Division :

4.4 Polynomial Long Division
Dividing : x3-4x2+x+26
("Dividend")
By : x+2 ("Divisor")

dividend x3 - 4x2 + x + 26
- divisor * x2 x3 + 2x2
remainder - 6x2 + x + 26
- divisor * -6x1 - 6x2 - 12x
remainder 13x + 26
- divisor * 13x0 13x + 26
remainder 0
Quotient : x2-6x+13 Remainder: 0

Trying to factor by splitting the middle term

4.5 Factoring x2-6x+13

The first term is, x2 its coefficient is 1 .
The middle term is, -6x its coefficient is -6 .
The last term, "the constant", is +13

Step-1 : Multiply the coefficient of the first term by the constant 1 • 13 = 13

Step-2 : Find two factors of 13 whose sum equals the coefficient of the middle term, which is -6 .

-13 + -1 = -14
-1 + -13 = -14
1 + 13 = 14
13 + 1 = 14

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

Equation at the end of step 4 :

(x2 - 6x + 13) • (x + 2) • (x - 3) = 0
Step 5 :

Solve (x2-6x+13)•(x+2)•(x-3) = 0
Theory - Roots of a product :

5.1 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 :

5.2 Find the Vertex of y = x2-6x+13

Parabolas 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 + 13.0
or y = 4.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2-6x+13
Axis of Symmetry (dashed) {x}={ 3.00}
Vertex at {x,y} = { 3.00, 4.00}
Function has no real roots


Solve Quadratic Equation by Completing The Square

5.3 Solving x2-6x+13 = 0 by Completing The Square .

Subtract 13 from both side of the equation :
x2-6x = -13

Now 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 :
-13 + 9 or, (-13/1)+(9/1)
The common denominator of the two fractions is 1 Adding (-13/1)+(9/1) gives -4/1
So adding to both sides we finally get :
x2-6x+9 = -4

Adding 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 = -4 and
x2-6x+9 = (x-3)2
then, according to the law of transitivity,
(x-3)2 = -4

We'll refer to this Equation as Eq. #5.3.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
(x-3)2 is
(x-3)2/2 =
(x-3)1 =
x-3

Now, applying the Square Root Principle to Eq. #5.3.1 we get:
x-3 = √ -4

Add 3 to both sides to obtain:
x = 3 + √ -4
In Math, i is called the imaginary unit. It satisfies i2 =-1. Both i and -i are the square roots of -1


Since a square root has two values, one positive and the other negative
x2 - 6x + 13 = 0
has two solutions:
x = 3 + √ 4 • i
or
x = 3 - √ 4 • i


Solve Quadratic Equation using the Quadratic Formula

5.4 Solving x2-6x+13 = 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 = 13

Accordingly, B2 - 4AC =
36 - 52 =
-16

Applying the quadratic formula :

6 ± √ -16
x = —————
2

In the set of real numbers, negative numbers do not have square roots. 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 1

Accordingly,√ -16 =
√ 16 • (-1) =
√ 16 • √ -1 =
± √ 16 • i

Can √ 16 be simplified ?

Yes! The prime factorization of 16 is
2•2•2•2
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 16 = √ 2•2•2•2 =2•2•√ 1 =
± 4 • √ 1 =
± 4

So now we are looking at:
x = ( 6 ± 4i ) / 2

Two imaginary solutions :

x =(6+√-16)/2=3+2i= 3.0000+2.0000i
or:
x =(6-√-16)/2=3-2i= 3.0000-2.0000i

Solving a Single Variable Equation :

5.5 Solve : x+2 = 0

Subtract 2 from both sides of the equation :
x = -2

Solving a Single Variable Equation :

5.6 Solve : x-3 = 0

Add 3 to both sides of the equation :
x = 3

Four solutions were found :

x = 3
x = -2
x =(6-√-16)/2=3-2i= 3.0000-2.0000i
x =(6+√-16)/2=3+2i= 3.0000+2.0000i

Answer 2

no offense but I didn't ask tiger algebra for help.

Answer 3

the other 2 zeros are \(x= -\dfrac{3}{2}\) and \(x=-1\)

Answer 4

well it's multiple choice and those don't go with my answer choices.

Answer 5

post the choices

Answer 6

they are;
a) positive: 4, 2 or 0; negative: 2 or 0; complex: 6, 4, 2 or 0
b) positive: 3 or 1; negative: 3 or 1; complex: 4, 2 or 0
c) Positive: 2 or 0; negative: 2 or 0; complex: 6, 4 or 2
d) Positive: 5, 3 or 1; negative: 1; complex: 4, 2 or 0