(4) Find factors of 6y²-5y-6
Answers
((2•3y2) - 5y) - 6 = 0
STEP
2
:
Trying to factor by splitting the middle term
2.1 Factoring 6y2-5y-6
The first term is, 6y2 its coefficient is 6 .
The middle term is, -5y its coefficient is -5 .
The last term, "the constant", is -6
Step-1 : Multiply the coefficient of the first term by the constant 6 • -6 = -36
Step-2 : Find two factors of -36 whose sum equals the coefficient of the middle term, which is -5 .
-36 + 1 = -35
-18 + 2 = -16
-12 + 3 = -9
-9 + 4 = -5 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -9 and 4
6y2 - 9y + 4y - 6
Step-4 : Add up the first 2 terms, pulling out like factors :
3y • (2y-3)
Add up the last 2 terms, pulling out common factors :
2 • (2y-3)
Step-5 : Add up the four terms of step 4 :
(3y+2) • (2y-3)
Which is the desired factorization
Equation at the end of step
2
:
(2y - 3) • (3y + 2) = 0
STEP
3
:
Theory - Roots of a product
3.1 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.
Solving a Single Variable Equation:
3.2 Solve : 2y-3 = 0
Add 3 to both sides of the equation :
2y = 3
Divide both sides of the equation by 2:
y = 3/2 = 1.500
Solving a Single Variable Equation:
3.3 Solve : 3y+2 = 0
Subtract 2 from both sides of the equation :
3y = -2
Divide both sides of the equation by 3:
y = -2/3 = -0.667
Supplement : Solving Quadratic Equation Directly
Solving 6y2-5y-6 = 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:
4.1 Find the Vertex of t = 6y2-5y-6
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 "t" because the coefficient of the first term, 6 , 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,Ay2+By+C,the y -coordinate of the vertex is given by -B/(2A) . In our case the y coordinate is 0.4167
Plugging into the parabola formula 0.4167 for y we can calculate the t -coordinate :
t = 6.0 * 0.42 * 0.42 - 5.0 * 0.42 - 6.0
or t = -7.042
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : t = 6y2-5y-6
Axis of Symmetry (dashed) {y}={ 0.42}
Vertex at {y,t} = { 0.42,-7.04}
y -Intercepts (Roots) :
Root 1 at {y,t} = {-0.67, 0.00}
Root 2 at {y,t} = { 1.50, 0.00}