Core Connections Algebra 1, 2013
CC
Core Connections Algebra 1, 2013 View details
2. Section 10.2
Continue to next subchapter

Exercise 69 Page 493

Practice makes perfect
a We want to factor the given polynomial. We can start by finding the greatest common factor (GCF) of the terms in the given expression. To do so we will consider coefficients and variables separately.
4 x^2- 12 x

Let's start by finding the GCF of 4 and 12. Factors of4:& 1,2,and 4 Factors of12:& 1,2,3, 4,6,and 12 We found that the GCF of the coefficients is 4. To find the GCF of the variables, we need to identify the variables repeated in both terms and write them with their minimum exponents. \begin{aligned} \textbf{Factors of }\bm{1^\text{st}}\textbf{ Variable:}&\ {\color{#FF0000}{x}}, x^2\\ \textbf{Factors of }\bm{2^\text{nd}}\textbf{ Variable:}&\ {\color{#FF0000}{x}} \end{aligned} We see that there is one repeated variable factor, x. Thus, the GCF of the expression is 4* x= 4x. Now we can write the given expression in terms of the GCF. 4x^2-12x ⇔ 4x* x- 4x* 3 Finally, we will factor out the GCF. 4x* x- 4x* 3 ⇔ 4x(x-3)

b Let's start factoring by first identifying the GCF. Then we will rewrite the expression as a trinomial with a leading coefficient of 1.

Factor Out the GCF

The GCF of an expression is a common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. In this case, the GCF is 3.
3y^2+6y+3
3* y^2+ 3* 2y+ 3* 1
3(y^2+2y+1)

The result of factoring out a GCF from the given expression is a trinomial with a leading coefficient of 1. 3(y^2+2y+1) Let's temporarily only focus on this trinomial, and we will bring back the GCF after factoring.

Factor the Expression

We want to factor the above expression. Factoring is much easier when our polynomial is a perfect square trinomial. To determine if an expression is a perfect square trinomial, we need to ask ourselves three questions.

Is the first term a perfect square? y^2= y^2 âś“
Is the last term a perfect square? 1= 1^2 âś“
Is the middle term twice the product of 1 and y? 2y=2* 1* y âś“

As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note there is an addition sign in the middle. y^2+2y+1 ⇔ ( y+ 1)^2 Wait! Before we finish, remember that we factored out a GCF from the original expression. To fully complete the factored expression, let's reintroduce that GCF now. 3(y+1)^2

c We want to completely factor the given expression. To do so, we will first identify and factor out the greatest common factor.

Factor Out the GCF

The GCF of an expression is the common factor of the terms in the expression. It is the common factor with the greatest coefficient and the greatest exponent. The GCF of the given expression is m.
2m^3+7m^2+3m
m(2m^2)+ m(7m)+ m(3)
m(2m^2+7x+3)

Factor the Quadratic Trinomial

Here we have a quadratic trinomial of the form am^2+bm+c, where |a| ≠ 1 and there are no common factors. To factor this expression, we will rewrite the middle term, bm, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b. m( 2m^2+7m+3 ) We have that a= 2, b=7, and c=3. There are now three steps we need to follow in order to rewrite the above expression.

  1. Find a c. Since we have that a= 2 and c=3, the value of a c is 2* 3=6.
  2. Find factors of a c. Since ac=6, which is positive, we need factors of a c to have the same sign — both positive or both negative — in order for the product to be positive. Since b=7, which is also positive, those factors will need to be positive so that their sum is positive.

c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result 1 & 6 & 1 + 6 &7 2 &3 &2 + 3 &5

  1. Rewrite bm as two terms. Now that we know which factors are the ones to be used, we can rewrite bm as two terms. m(2m^2+7m+3 ) ⇕ m ( 2m^2+ 1m+ 6m+3 )
Finally, we will factor the last expression obtained.
m( 2m^2+m+6m+3 )
m( m(2m+1)+6m+3 )
m( m(2m+1)+3(2m+1) )
m ( (m+3)(2m+1) )
m(m+3)(2m+1)
d Here we have a quadratic trinomial of the form ax^2+bx+c, where |a| ≠ 1 and there are no common factors. To factor this expression we will rewrite the middle term, bx, as two terms. The coefficients of these two terms will be factors of ac whose sum must be b.
3x^2+4x-4 ⇔ 3x^2+4x+(- 4) We have that a= 3, b=4, and c=- 4. There are now three steps we need to follow in order to rewrite the above expression.
  1. Find a c. Since we have that a= 3 and c=- 4, the value of a c is 3* (- 4)=- 12.
  2. Find factors of a c. Since ac=- 12, which is negative, we need factors of a c to have opposite signs — one positive and one negative. Since b=4, which is positive, the absolute value of the positive factor will need to be greater than the absolute value of the negative factor, so that their sum is positive.

c|c|c|c 1^(st)Factor &2^(nd)Factor &Sum &Result - 1 &12 &- 1 + 12 &11 - 2 & 6 & - 2 + 6 &4 - 3 &4 &- 3 + 4 &1

  1. Rewrite bx as two terms. Now that we know which factors are the ones to be used, we can rewrite bx as two terms. 3x^2+4x-4 ⇕ 3x^2 - 2x + 6x-4
Finally, we will factor the last expression obtained.
3x^2-2x+6x-4
x(3x-2)+6x-4
x(3x-2)+2(3x-2)
(x+2)(3x-2)