Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1Factoring z2-10z+25
The first term is, z2 its coefficient is 1.
The middle term is, -10z its coefficient is -10.
The last term, "the constant", is +25
Step-1 : Multiply the coefficient of the first term by the constant 1•25=25
Step-2 : Find two factors of 25 whose sum equals the coefficient of the middle term, which is -10.
| -25 | + | -1 | = | -26 | ||
| -5 | + | -5 | = | -10 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step2above, -5 and -5
z2 - 5z-5z - 25
Step-4 : Add up the first 2 terms, pulling out like factors:
z•(z-5)
Add up the last 2 terms, pulling out common factors:
5•(z-5)
Step-5:Add up the four terms of step4:
(z-5)•(z-5)
Which is the desired factorization
Multiplying Exponential Expressions:
1.2 Multiply (z-5) by (z-5)
The rule says : To multiply exponential expressions which have the same base, add up their exponents.
In our case, the common base is (z-5) and the exponents are:
1,as(z-5) is the same number as (z-5)1
and1,as(z-5) is the same number as (z-5)1
The product is therefore, (z-5)(1+1) = (z-5)2
Equation at the end of step 1 :
(z - 5)2 = 0 Step 2 :
Solving a Single Variable Equation:
2.1Solve:(z-5)2 = 0(z-5)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: z-5=0
Add 5 to both sides of the equation:
z = 5
Supplement : Solving Quadratic Equation Directly
Solving z2-10z+25 = 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:
3.1Find the Vertex ofy = z2-10z+25Parabolas 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,Az2+Bz+C,the z-coordinate of the vertex is given by -B/(2A). In our case the z coordinate is 5.0000Plugging into the parabola formula 5.0000 for z we can calculate the y-coordinate:
y = 1.0 * 5.00 * 5.00 - 10.0 * 5.00 + 25.0
or y = 0.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = z2-10z+25
Vertex at {z,y} = { 5.00, 0.00}
z-Intercept (Root) :
One Root at {z,y}={ 5.00, 0.00}
Note that the root coincides with
the Vertex and the Axis of Symmetry
coinsides with the line z = 0
Solve Quadratic Equation by Completing The Square
3.2Solvingz2-10z+25 = 0 by Completing The Square.Subtract 25 from both side of the equation :
z2-10z = -25
Now the clever bit: Take the coefficient of z, which is 10, divide by two, giving 5, and finally square it giving 25
Add 25 to both sides of the equation :
On the right hand side we have:
-25+25or, (-25/1)+(25/1)
The common denominator of the two fractions is 1Adding (-25/1)+(25/1) gives 0/1
So adding to both sides we finally get:
z2-10z+25 = 0
Adding 25 has completed the left hand side into a perfect square :
z2-10z+25=
(z-5)•(z-5)=
(z-5)2
Things which are equal to the same thing are also equal to one another. Since
z2-10z+25 = 0 and
z2-10z+25 = (z-5)2
then, according to the law of transitivity,
(z-5)2 = 0
We'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
(z-5)2 is
(z-5)2/2=
(z-5)1=
z-5
Now, applying the Square Root Principle to Eq.#3.2.1 we get:
z-5= √ 0
Add 5 to both sides to obtain:
z = 5 + √ 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:
z = 5
Solve Quadratic Equation using the Quadratic Formula
3.3Solvingz2-10z+25 = 0 by the Quadratic Formula.According to the Quadratic Formula,z, the solution forAz2+Bz+C= 0 , where A, B and C are numbers, often called coefficients, is given by :
-B± √B2-4AC
z = ————————
2A In our case,A= 1
B=-10
C= 25 Accordingly,B2-4AC=
100 - 100 =
0Applying the quadratic formula :
10 ± √ 0
z=————
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:
z = 10 / 2 = 5
One solution was found :
z = 5
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Now, let's dive into the concepts used in this article:
Factoring a Quadratic Polynomial:
The article discusses the process of factoring a quadratic polynomial, specifically the expression z^2 - 10z + 25. The goal is to factor this expression into its simplest form.
The article suggests using the method of "splitting the middle term" to factor the quadratic polynomial. This method involves finding two numbers whose product is equal to the product of the coefficient of the quadratic term and the constant term, and whose sum is equal to the coefficient of the linear term. In this case, the coefficient of the quadratic term is 1, the coefficient of the linear term is -10, and the constant term is 25.
By applying the method, the article shows that the two numbers that satisfy these conditions are -5 and -5. Therefore, the quadratic polynomial can be factored as (z - 5)(z - 5), which can be further simplified as (z - 5)^2.
Multiplying Exponential Expressions:
The article also discusses the process of multiplying exponential expressions. Specifically, it demonstrates the multiplication of (z - 5) by itself, which is written as (z - 5)^2.
According to the article, when multiplying exponential expressions with the same base, the exponents are added. In this case, the base is (z - 5), and the exponents are both 1. Therefore, the product is (z - 5)^(1+1) = (z - 5)^2.
Solving a Quadratic Equation:
The article then moves on to solving the quadratic equation (z - 5)^2 = 0. It presents two methods for solving this equation: completing the square and using the quadratic formula.
Completing the square involves manipulating the equation to express it in the form (z - p)^2 = q, where p and q are constants. In this case, the equation is already in this form, with p = 5 and q = 0. Therefore, the solutions to the equation are z = 5.
The quadratic formula is another method for solving quadratic equations. It states that for an equation in the form az^2 + bz + c = 0, the solutions are given by the formula z = (-b ± √(b^2 - 4ac)) / (2a). Applying this formula to the equation (z - 5)^2 = 0, we find that the solutions are z = 5.
Finding the Vertex of a Parabola:
The article briefly discusses finding the vertex of a parabola. It mentions that the vertex is the highest or lowest point of a parabola and that it can be used to determine various properties of the parabola.
In the specific example given in the article, the equation of the parabola is y = z^2 - 10z + 25. The article states that the vertex of this parabola can be found using the formula -B/(2A), where A and B are the coefficients of the quadratic and linear terms, respectively. In this case, A = 1 and B = -10, so the z-coordinate of the vertex is 5. By substituting this value into the equation, the y-coordinate of the vertex is found to be 0.
These are the main concepts discussed in this article. If you have any further questions or need clarification on any specific aspect, feel free to ask!
