Factoring

Common factor by grouping terms

A grouping of terms is a common factor polynomial

• To the factor, we must first know how to identify how we are going to form the factors. Here we try to get two factors or more in which by multiplying them, we obtain the original expression

• It begins with a step similar to the previous case, and it's to find what common factor can be extracted from a group of expressions

• It's a good idea to start by looking for the constants that are the same in a group of expressions to factor them out

• It's important that you pay attention to the signs, since depending on how they're organized in the original expression, they will change their order in their factored equivalent

• Factoring does the opposite of notable products

Monomial common factor

Factor: It's what results from a product. For example, if we multiply X by 3, one factor is X and the other factor is 3. A factor can be either a variable or a constant

Common: All the terms of our monomial have this factor. We can extract this common factor from all of them

Monomial: It's a single term with its respective sign

Factorization: It's to make our expression smaller

• When factoring with a common monomial factor, we must look for which terms are repeated in each of the constants or variables of our monomial

• Our constants or numbers that accompany the letters or variables must also be factored out. To do that, all expressions must be found to be divisible by a number. That number is the factor to remove

Perfect square trinomial (PST)

Earlier we learned how to solve the binomial square. Now we're going to go backward, factoring back to the squared binomial

To perform this factorization:

• We must verify that it's a perfect square trinomial:

  1. We must have 3 terms, that's, a trinomial

  2. Our first term and our third (ordering our trinomial) have perfect square roots, the variables being raised to even exponents

  3. We multiply the value of the root of the first and third terms, all of that multiplied by two. The result has to give us the same as the value of the second term of our trinomial

Once we gain experience taking the roots without the need for algebraic operations, it's enough to extract the roots, extract the sign, and square everything

Difference of perfect squares

When an expression does not yield or it's difficult for us to solve it, we've to search and identify what type of expression it's, to choose the appropriate factorization case

• We can easily identify that it's a difference of perfect squares because the terms are going to be subtracted and each one is squared (so that the square root can be taken)

• It's solved by finding the roots, opening and closing a pair of parentheses, and placing the roots in both; the order must be as it appears in our original expression. One of the parentheses will have a plus and another a minus

• By factoring, we will have conjugate binomials. Binomials because they're two terms and conjugates because one of the terms has a positive sign and the other has a negative sign

PST by addition and subtraction

PST by addition and subtraction is also known as the method of completing the squares. This method is very similar to the perfect square trinomial with some differences in its second term

• We must verify that we've a trinomial

• The first and second terms must have a perfect square root

• It's in the second term that this method differs from that of the perfect square trinomial, since its value is not equal to the first and third terms multiplied by 2

• We solve the system as if it were a perfect square trinomial, and what we need to have the complete second term (that complies with the laws of the perfect square trinomial) we're going to add it and in order not to lose the balance in the system, we're going to subtract as well. What we put we've to remove

• We factor normally as a perfect square trinomial, leaving out of the resulting parentheses the part that we subtracted in earlier steps when balancing the system

• In many cases the resulting expression will be a difference of perfect squares, which we can factor again

• It's advisable to always leave the system in its minimum expression

Trinomial of the form x^2+bx+c

For this factorization case, we must be very careful that our first term (which must be squared) isn't accompanied by a constant or number; additionally, the second term must have a number accompanying the variable of the first term but without being squared; and the third term must be a single constant

• We must open two factors or parentheses. The first term of each parenthesis will be the square root of the first term of our original expression

• We must look for factors of our third term of the original expression; that's, numbers that multiplied together give us the value of that term, taking into account that if the term is negative, one of the factors must be positive and the other negative, and if it's positive, both factors must be positive or negative

• Once we've got the factors, we must select the two that, added or subtracted, give us the value of the constant that's in the second term of the original expression. Those two factors will be the ones that go in the second terms of each of the two parentheses

Material: Factorization

Case 1: Monomial common factor

When all terms of a polynomial have a common factor

1. Decompose into factors

$x^2+2x=x(x+2)$

1. Decompose into factors

$5y-15xy^2=5y(1-3xy)$

1. Decompose into factors

$$\begin{split} 5z^2-5z+15z^3 & =5z(5z-1+3z^2)\\ & =5z(3z^2+5z-1) \end{split}$$

1. Decompose into factors

$$\begin{split} 9my^2-27m^2x^2y^2+18my^2 & =27my^2-27m^2x^2y^2\\ & =27my^2(1-mx^2)\\ & =27my^2(-mx^2+1) \end{split}$$

1. Decompose into factors

$$\begin{split} 6xy^3-9mx^2y^3+12nx^3y^3-3n^2x^4y^3 & =-3n^2x^4y^3+12nx^3y^3-9mx^2y^3+6xy^3\\ & =3xy^3(-n^2x^3+4nx^2-3mx+2) \end{split}$$

Case 2: Common factor by grouping terms

1. Decompose

$$\begin{split} ax+bx+ay+by & =(ax+bx)+(ay+by)\\ & =x(a+b)+y(a+b)\\ & =(x+y)(a+b) \end{split}$$

1. Decompose

$$\begin{split} am-bm+an-bn & =(am-bm)+(an-bn)\\ & =m(a-b)+n(a-b)\\ & =(m+n)(a-b) \end{split}$$

1. Decompose

$$\begin{split} 6ax+3a+1+2x & =6ax+3a+2x+1\\ & =3a(2x+1)+(2x+1)\\ & =(2x+1)(3a+1) \end{split}$$

1. Decompose

$$\begin{split} 3x^2-9ax-x+3a & =3x^2-x-9ax+3a\\ & =(3x^2-x)-(9ax+3a)\\ & =(3x^2-x)+(-9ax-3a)\\ & =x(3x-1)+3a(3x-1)\\ & = (x+3a)(3x-1) \end{split}$$

1. Decompose

$$\begin{split} 2a^2x-5a^2y+15by-6bx & =(2a^2x-5a^2y)+(15by-6bx)\\ & =a^2(2x-5y)+3b(5y-2x)\\ & =a^2(2x-5y)+3b(-2x+5y)\\ & =a^2(2x-5y)-3b(2x-5y)\\ & =(a^2-3b)(2x-5y) \end{split}$$

Case 3: Perfect square trinomial

A quantity is a perfect square when it's the square of another quantity. That's when it's the product of two equal factors

1. Decompose

$1-14x^2y+49x^4y^2$ $\sqrt{1} = 1; (\sqrt{1})^2=1^2; 1=1$ $\sqrt{49x^4y^2}=7x^2y; (\sqrt{49x^4y^2})^2=(7x^2y)^2;49x^4y^2=(7x^2y)^2$

$$\begin{split} 1-14x^2y+49x^4y^2 &=1-14x^2y+(7x^2y)^2\\ & =(1+7x^2y)^2 \end{split}$$