Variables - PSAT Math

Let's try to factor x2 – y2 – z2 + 2yz.

Notice that the last three terms are very close to y2 + z2 – 2yz, which, if we rearranged them, would become y2 – 2yz+ z2. We could factor y2 – 2yz+ z2 as (y – z)2, using the general rule that p2 – 2pq + q2 = (p – q)2 .

So we want to rearrange the last three terms. Let's group them together first.

x2 + (–y2 – z2 + 2yz)

If we were to factor out a –1 from the last three terms, we would have the following:

x2 – (y2 + z2 – 2yz)

Now we can replace y2 + z2 – 2yz with (y – z)2.

x2 – (y – z)2

This expression is actually a differences of squares. In general, we can factor p2 – q2 as (p – q)(p + q). In this case, we can substitute x for p and (y – z) for q.

x2 – (y – z)2 = (x – (y – z))(x + (y – z))

Now, let's distribute the negative one in the trinomial x – (y – z)

(x – (y – z))(x + (y – z))

(x – y + z)(x + y – z)

The problem said that factoring x2 – y2 – z2 + 2yz would result in two polynomials in the form (ax + by + cz)(dx + ey + fz), where a, b, c, d, e, and f were all integers, and a > 0.

(x – y + z)(x + y – z) fits this form. This means that a = 1, b = –1, c = 1, d = 1, e = 1, and f = –1. The sum of all of these is 2.

The answer is 2.

Nothing common cancels at the beginning. To factor this, we need to find two numbers that multiply to 9 * 4 = 36 and sum to 12. 6 and 6 work.

So 9_x_2 + 12_x_ + 4 = 9_x_2 + 6_x_ + 6_x_ + 4

Let's look at the first two terms and last two terms separately to begin with. 9_x_2 + 6_x_ can be simplified to 3_x_(3_x_ + 2) and 6_x_ + 4 can be simplified into 2(3_x_ + 2). Putting these together gets us

9_x_2 + 12_x_ + 4

= 9_x_2 + 6_x_ + 6_x_ + 4

= 3_x_(3_x_ + 2) + 2(3_x_ + 2)

= (3_x_ + 2)(3_x_ + 2)

This is as far as we can factor.

$64y^{2}$ - 16 is a difference of squares.

The difference of squares formula is $a^{2}$ - $b^{2}$ = (a - b)(a + b).

Therefore, $$\frac{64y^{2}$$ - 16}{8y + 4} = $\frac{(8y + 4)(8y - 4)}{8y + 4}$ = 8y - 4.

The numerator on the left can be factored so the expression becomes $\frac{left ( x+3 right )times left ( x-3 right )}{left ( x+3 right )}$=5, which can be simplified to left ( x-3 right )=5

Then you can solve for x by adding 3 to both sides of the equation, so x=8

We can first factor out -3:

$-3(4x^{2}$-9)

This factors further because there is a difference of squares:

-3(2x+3)(2x-3)

First, factor.

$x^2$+3x+2=(x+2)(x+1)=0

Set each factor equal to 0

x+2=0; x=-2

x+1=0; x=-1

Therefore,

x=-2 or-1

Use the distributive property:

Combine like terms:

Find the product:

Step 1: Use the Distributive Property

Step 2: Combine like terms

Find the product:

Use the distributive property:

Use the distributive property:

Combine like terms:

Find the product:

Step 1: Use the Distributive Property

Step 2: Combine like terms

Find the product:

Use the distributive property:

The degree of the polynomial is the largest degree of any one of it's individual terms.

The degree of is

The degree of is

The degree of is

The degree of is

The degree of is

is the largest degree of any one of the terms of the polynomial, and so the degree of the polynomial is .

The polynomial has one term, so its degree is the sum of the exponents of the variables:

The degree of a polynomial in more than one variable is the greatest degree of any of the terms; the degree of a term is the sum of the exponents. The degrees of the terms in the given polynomial are:

The degree of the polynomial is the greatest of these degrees, 100.

The degree of a polynomial in one variable is the greatest exponent of any of the powers of the variable. The terms have as their exponents, in order, 44, 20, 10, and 100; the greatest of these is 100, which is the degree.

The degree of a polynomial in one variable is the greatest exponent of any of the powers of the variable. The terms have as their exponents, in order, 10, 20, 30, and 40; 40 is the greatest of them and is the degree of the polynomial.

The degree of a polynomial is the highest degree of any term; the degree of a term is the exponent of its variable or the sum of the exponents of its variables, with unwritten exponents being equal to 1. For each term in a polynomial, write the exponent or add the exponents; the greatest number is its degree. We do this with all four choices:

:

A constant term has degree 0.

The degree of this polynomial is 5.

The degree of this polynomial is 5.

The degree of this polynomial is 5.

The degree of this polynomial is 5.

All four polynomials have the same degree.

The degree of a monomial term is the sum of the exponents of its variables, with the default being 1.

For each monomial, this sum - and the degree - is as follows:

:

:

: (note - 999 is the coefficient)

:

is the correct choice.