## Step by step solution :

## Step 1 :

Trying to factor by splitting the middle term1.1Factoring x2-15x-100 The first term is, x2 its coefficient is 1.The middle term is, -15x its coefficient is -15.The last term, “the constant”, is -100Step-1 : Multiply the coefficient of the first term by the constant 1•-100=-100Step-2 : Find two factors of -100 whose sum equals the coefficient of the middle term, which is -15.

-100 | + | 1 | = | -99 | ||

-50 | + | 2 | = | -48 | ||

-25 | + | 4 | = | -21 | ||

-20 | + | 5 | = | -15 | That”s it |

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step2above, -20 and 5×2 – 20x+5x – 100Step-4 : Add up the first 2 terms, pulling out like factors:x•(x-20) Add up the last 2 terms, pulling out common factors:5•(x-20) Step-5:Add up the four terms of step4:(x+5)•(x-20)Which is the desired factorization

Equation at the end of step 1 :

(x + 5) • (x – 20) = 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+5 = 0Subtract 5 from both sides of the equation:x = -5

Solving a Single Variable Equation:2.3Solve:x-20 = 0Add 20 to both sides of the equation:x = 20

### Supplement : Solving Quadratic Equation Directly

Solving x2-15x-100 = 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 = x2-15x-100Parabolas 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.

You are watching: Use the zero product property to find the solutions to the equation x2 – 15x – 100 = 0.

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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 7.5000Plugging into the parabola formula 7.5000 for x we can calculate the y-coordinate:y = 1.0 * 7.50 * 7.50 – 15.0 * 7.50 – 100.0 or y = -156.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for : y = x2-15x-100 Axis of Symmetry (dashed) {x}={ 7.50} Vertex at {x,y} = { 7.50,-156.25} x-Intercepts (Roots) : Root 1 at {x,y} = {-5.00, 0.00} Root 2 at {x,y} = {20.00, 0.00}

Solve Quadratic Equation by Completing The Square

3.2Solvingx2-15x-100 = 0 by Completing The Square.Add 100 to both side of the equation : x2-15x = 100Now the clever bit: Take the coefficient of x, which is 15, divide by two, giving 15/2, and finally square it giving 225/4Add 225/4 to both sides of the equation :On the right hand side we have:100+225/4or, (100/1)+(225/4)The common denominator of the two fractions is 4Adding (400/4)+(225/4) gives 625/4So adding to both sides we finally get:x2-15x+(225/4) = 625/4Adding 225/4 has completed the left hand side into a perfect square :x2-15x+(225/4)=(x-(15/2))•(x-(15/2))=(x-(15/2))2 Things which are equal to the same thing are also equal to one another. Sincex2-15x+(225/4) = 625/4 andx2-15x+(225/4) = (x-(15/2))2 then, according to the law of transitivity,(x-(15/2))2 = 625/4We”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(x-(15/2))2 is(x-(15/2))2/2=(x-(15/2))1=x-(15/2)Now, applying the Square Root Principle to Eq.#3.2.1 we get:x-(15/2)= √ 625/4 Add 15/2 to both sides to obtain:x = 15/2 + √ 625/4 Since a square root has two values, one positive and the other negativex2 – 15x – 100 = 0has two solutions:x = 15/2 + √ 625/4 orx = 15/2 – √ 625/4 Note that √ 625/4 can be written as√625 / √4which is 25 / 2

### Solve Quadratic Equation using the Quadratic Formula

3.3Solvingx2-15x-100 = 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=-15C=-100 Accordingly,B2-4AC=225 – (-400) = 625Applying the quadratic formula : 15 ± √ 625 x=——————2Can √ 625 be simplified ?Yes!The prime factorization of 625is5•5•5•5 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).√ 625 =√5•5•5•5 =5•5•√ 1 =±25 •√ 1 =±25 So now we are looking at:x=(15±25)/2Two real solutions:x =(15+√625)/2=(15+25)/2= 20.000 or:x =(15-√625)/2=(15-25)/2= -5.000

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