Equations / Solution Sets - PSAT Math

In general, if a function has a root at x = r, then (x – r) must be a factor of f(x). In this problem, we are told that f(x) has roots at –1, 0 and 2. This means that the following are all factors of f(x):

(x – (–1)) = x + 1

(x – 0) = x

and (x – 2).

This means that we must look for an equation for f(x) that has the factors (x + 1), x, and (x – 2).

We can immediately eliminate the function f(x) = _x_2 + x – 2, because we cannot factor an x out of this polynomial. For the same reason, we can eliminate f(x) = _x_2 – x – 2.

Let's look at the function f(x) = _x_3 – x_2 + 2_x. When we factor this, we are left with x(_x_2 – x + 2). We cannot factor this polynomial any further. Thus, x + 1 and x – 2 are not factors of this function, so it can't be the answer.

Next, let's examine f(x) = _x_4 + _x_3 – 2_x_2 .

We can factor out _x_2.

_x_2 (_x_2 + x – 2)

When we factor _x_2 + x – 2, we will get (x + 2)(x – 1). These factors are not the same as x – 2 and x + 1.

The only function with the right factors is f(x) = _x_3 – x_2 – 2_x.

When we factor out an x, we get (_x_2 – x – 2), which then factors into (x – 2)(x + 1). Thus, this function has all of the factors we need.

The answer is f(x) = _x_3 – x_2 – 2_x.

The numbers by which _x_6 – _y_6 is divisible will be all of its factors. In other words, we need to find all of the factors of _x_6 – _y_6 , which essentially means we must factor _x_6 – _y_6 as much as we can.

First, we will want to apply the difference of squares rule, which states that, in general, _a_2 – _b_2 = (a – b)(a + b). Notice that a and b are the square roots of the values of _a_2 and _b_2, because √_a_2 = a, and √_b_2 = b (assuming a and b are positive). In other words, we can apply the difference of squares formula to _x_6 – _y_6 if we simply find the square roots of _x_6 and _y_6.

Remember that taking the square root of a quantity is the same as raising it to the one-half power. Remember also that, in general, (ab)c = abc.

√_x_6 = (_x_6)(1/2) = x(6(1/2)) = _x_3

Similarly, √_y_6 = _y_3.

Let's now apply the difference of squares factoring rule.

_x_6 – _y_6 = (_x_3 – _y_3)(_x_3 + _y_3)

Because we can express _x_6 – _y_6 as the product of (_x_3 – _y_3) and (_x_3 + _y_3), both (_x_3 – _y_3) and (_x_3 + _y_3) are factors of _x_6 – _y_6 . Thus, we can eliminate _x_3 – _y_3 from the answer choices.

Let's continue to factor (_x_3 – _y_3)(_x_3 + _y_3). We must now apply the sum of cubes and differences of cubes formulas, which are given below:

In general, _a_3 + _b_3 = (a + b)(_a_2 – ab + _b_2). Also, _a_3 – _b_3 = (a – b)(_a_2 + ab + _b_2)

Thus, we have the following:

(_x_3 – _y_3)(_x_3 + _y_3) = (x – y)(_x_2 + xy + _y_2)(x + y)(_x_2 – xy + _y_2)

This means that x – y and x + y are both factors of _x_6 – _y_6 , so we can eliminate both of those answer choices.

We can rearrange the factorization (x – y)(_x_2 + xy + _y_2)(x + y)(_x_2 – xy + _y_2) as follows:

(x – y)(x + y)(_x_2 + xy + _y_2)(_x_2 – xy + _y_2)

Notice that (x – y)(x + y) is merely the factorization of difference of squares. Therefore, (x – y)(x + y) = _x_2 – _y_2.

(x – y)(x + y)(_x_2 + xy +_y_2)(_x_2 – xy + _y_2) = (_x_2 – _y_2)(_x_2 + xy +_y_2)(_x_2 – xy + _y_2)

This means that _x_2 – _y_2 is also a factor of _x_6 – _y_6.

By process of elimination, _x_2 + _y_2 is not necessarily a factor of _x_6 – _y_6 .

The answer is _x_2 + _y_2 .

First pull out any common terms: 4_x_3 – 16_x_ = 4_x_(_x_2 – 4)

_x_2 – 4 is a difference of squares, so we can also factor that further. The difference of squares formula is _a_2 – _b_2 = (a – b)(a + b). Here a = x and b = 2. So _x_2 – 4 = (x – 2)(x + 2).

Putting everything together, 4_x_3 – 16_x_ = 4_x_(x + 2)(x – 2).

This is a difference of squares. The difference of squares formula is a_2 – b_2 = (a + b)(a – b). In this problem, a = 6_x and b = 7_y.

So 36_x_2 – 49_y_2 = (6_x_ + 7_y_)(6_x_ – 7_y_).

Find two numbers that add to and multiply to

Factors of

You can use

Then make each factor equal 0.

and

and

Find all factors of 24

1, 2, 3,4, 6, 8, 12, 24

Now find two factors that add up to and multiply to ; and are the two factors.

By factoring, you can set the equation to be

If you FOIL it out, it gives you .

Set each part of the equation equal to 0, and solve for .

and

and

Factoring yields giving roots of and .

The numerator can be factored into .

Therefore, it can cancel with the denominator. So imples .

We can factor out a , leaving .

From there we can factor again to

.

First pull out any common terms: 4_x_3 – 16_x_ = 4_x_(_x_2 – 4)

_x_2 – 4 is a difference of squares, so we can also factor that further. The difference of squares formula is _a_2 – _b_2 = (a – b)(a + b). Here a = x and b = 2. So _x_2 – 4 = (x – 2)(x + 2).

Putting everything together, 4_x_3 – 16_x_ = 4_x_(x + 2)(x – 2).

Start by looking at your last term. Since this term is negative, you will need to have a positive group and a negative group:

Now, since the middle term is positive, you can guess that the positive group will contain the larger number. Likewise, since the coefficient is only , you can guess that the factors will be close. Two such factors of are and .

Therefore, your groups will be:

Begin by looking at the last element. Since it is positive, you know that your groups will contain either two additions or two subtractions. Since the middle term is negative (), your groups will be two subtractions:

Now, the factors of are and , and , and and .

Clearly, the last is the one that works, for when you FOIL , you get your original equation!

The original statement is equivalent to a statement that is identically true regardless of the value of ; therefore, so is the original statement itself. The solution set is the set of all real numbers.

To solve this equation, simply plug 12 in for in the equation.

Factor the polynomial as the difference of squares:

We set each binomial equal to 0 and apply the Square Root Property:

This yields two imaginary solutions.

This yields two irrational solutions.

The correct response is that the solution set comprises two irrational numbers and two imaginary numbers.

The question is equivalent to asking the following:

For what value of does the equation

have more than one solution?

The equation simplifies as follows:

If the absolute values of two expressions equal, then either the expressions themselves are equal or they are each other's opposite.

Taking the latter case:

Regardless of the value of , exactly one solution is yielded this way.

The question becomes as follows: for which value of does the other way yield a solution?

Set:

If this is a false statement, then this yields no solutions.

If this is a true statement, then this automatically yields the set of all real numbers as the solution set. We solve for :

As a result, the statement

has infinitely many solutions, and the other three statements have exactly one.

First, since the equation is quadratic, put it in standard form

as follows:

To determine the nature of the solution set, evaluate discriminant for :

The discriminant is positive, but not a perfect square, so the solution set comprises two irrational numbers.

First, since the equation is quadratic, put it in standard form

as follows:

To determine the nature of the solution set, evaluate discriminant for :

The discriminant is positive and a perfect square, so the solution set comprises two rational numbers.

First, since the equation is quadratic, put it in standard form

as follows:

To determine the nature of the solution set, evaluate discriminant for :

The discriminant is negative, so the solution set comprises two imaginary numbers.

The cubic binomial in the equation can be factored to yield an equivalent equation as follows::

One of these factors must be equal to 0.

If then , so one solution is rational.

If , we can find out about the nature of the remaining solutions using discriminant , setting :

The discriminant is negative, so the two solutions of the equation are imaginary. These are also the two remaining solutions of .

The correct response is that the equation has one rational solution and two imaginary solutions.