You factor the equation and get:

(x+ 2)(x+ 1) = 0

Then, you getx= –2 andx= –1.

There are two ways to solve this problem. One option is to use the quadratic formula:

In this problem, a = 1, b = 1, and c = –6. However, there is a lot of room for computational error using this method.

Another option is to factor the equation using the reverse FOIL method. Because we have x², we know that both factors must include x. Thus, so far we have (x )(x ). We also know that one factor must be negative and one must be positive if our last value (the L in FOIL) is to be negative. Thus, we have (x– ) (x+ ). The factors of 6 are 1, 2, 3, and 6, so our last values must multiply together to make 6. Therefore, our options are (x – 1)(x + 6), (x – 6)(x + 1), (x – 2)(x + 3), and (x – 3)(x + 2). However, because we need our middle value to be 1, the combination of the outer and inner values (O and I in FOIL), must add up to positive 1. Since only a combination of +/–2 and +/–3 can add to 1, we can eliminate our first two options. Then, because we need the positive value to be larger than the negative value, we can determine that the correct factorization is (x – 2)(x + 3). When we set each of these pieces equal to zero, we get x – 2 = 0 and x + 3 = 0. Thus, X equals 2 and –3. We can check our work using FOIL.

1.因素上面的方程:

2. Solve for:

and...

To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)

(x+ 2)(x+ 3) = 0

x= –2 orx= –3

To solve, begin by factoring the equation. We know that in our two factors, one will begin withand one will begin with. Fiddling with different factors of(we are looking for two numbers that, when multiplied byandseparately, will add to), we come to the following:

(If you are unsure, double check by expanding the equation to match the original)

Now, set the each factor equal to 0:

Do the same for the second factor:

Therefore, our two values ofareand.

Alternatively, this problem can be solved by plugging each answer choice into the original equation and finding which set of numbers make the equation equate to 0.

Given a quadratic equation equal to zero you can factor the equation and set each factor equal to zero. To factor you have to find two numbers that multiply to make 9 and add to make 6. The number is 3. So the factored form of the problem is (x+3)(x+3)=0. This statement is true only when x+3=0. Solving for x gives x=-3. Since this problem is multiple choice, you could also plug the given answers into the equation to see which one works.

64x²+ 24x - 10 = 0

Factor the equation:

(8x + 5)(8x – 2) = 0

Set each side equal to zero

(8x + 5) = 0

x = -5/8

(8x – 2) = 0

x = 2/8 = 1/4

The roots of a function are thexintercepts of the function. Whenever a function passes through a point on thex-axis, the value of the function is zero. In other words, to find the roots of a function, we must set the function equal to zero and solve for the possible values ofx.

This is a quadratic trinomial. Let's see if we can factor it. (We could use the quadratic formula, but it's easier to factor when we can.)

Because the coefficient in front of theis not equal to 1, we need to multiply this coefficient by the constant, which is –4. When we mutiply 2 and –4, we get –8. We must now think of two numbers that will multiply to give us –8, but will add to give us –7 (the coefficient in front of thexterm). Those two numbers which multiply to give –8 and add to give –7 are –8 and 1. We will now rewrite –7x as –8x + x.

We will then group the first two terms and the last two terms.

We will next factor out a 2xfrom the first two terms.

Thus, when factored, the original equation becomes (2x+ 1)(x –4) = 0.

We now set each factor equal to zero and solve forx.

Subtract 1 from both sides.

2x= –1

Divide both sides by 2.

Now, we setx– 4 equal to 0.

x– 4 = 0

Add 4 to both sides.

x= 4

The roots off(x) occur wherex=.

The answer is therefore.

3.6x²- 12x - 15 = 0

Factor the equation:

(6 x + 3) (6 x -5) = 0

Set each side equal to zero

6x + 3 = 0

x = -3/6 = -1/2

6x – 5 = 0

x = 5/6

With quadratic equations, always begin by getting it into standard form:

Therefore, take our equation:

And rewrite it as:

You could use the quadratic formula to solve this problem. However, it is possible to factor this if you are careful. Factored, the equation can be rewritten as:

Now, either one of the groups on the left could beand the whole equation would be. Therefore, you set up each as a separate equation and solve for:

OR

The sum of these values is: