Factorise and solve : x^3-x^2-4x+4
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Answer:
Step-by-step explanation:
I understand it! (But I'm a high school math teacher so it might be unfair!) ; )
Anyway you have 2 numbers. You don't know what they are so we will call them x and y. Our first statement says that these numbers multiply to equal 120. So that first equation is
xy = 120
Our second statement says that they add to -22. So that second equation is
x + y = -22
Now we need to find the 2 numbers. We have 2 unknowns, x and y. That means that we have to have 2 equations. No more than 2 no less than 2. Solve the first equation for y. It doesn't matter which equation you pick and which variable you choose to solve for. If you do the work correctly, you will get the correct answers either way. Solving the first equation for y gives you:
Now tht we have a value for y, we can use the substitution property of equality to sub that value into the second equation for y:
Instead of finding a common denominator, we are going to eliminate x in the denominator altogether. Do this by multiplying everything by x to get
Get everything on one side of the equals sign and factor:
Because 12 * 10 = 120 and 12 + 10 = 22, our factors are 12 and 10:
Use factoring by grouping now.
From the first set of parenthesis you can factor out an x, and from the second set you can factor out a 10:
x(x + 12) + 10(x + 12) = 0
Now factor out the common x + 12 to get a factored quadratic:
(x + 12)(x + 10) = 0
By the Zero Product Property, either x + 12 = 0 or x + 10 = 0. Solving both for x gives you x = -12 or x = -10.
If x = -12, then
-12 + y = -22 so
y = -10
If x = -10, then
-10 + y = -22 so
y = -12.
The combinations of x and y are:
If x = -12, y = 10
If x = -10, y = -12
It doesn't matter which one is which. They are not asking you to specifically define x and y. They are asking you to find the numbers, and you did!