Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 7.2
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Exercise 92 Page 408

Practice makes perfect
a To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.
x^2-8x+7 In this case we have 7. This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign — both positive or both negative.
Factor Constants Product of Constants
1 and 7 7
-1 and - 7 7

Next, let's consider the coefficient of the linear term. x^2- 8x+7 For this term we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 8.

Factors Sum of Factors
1 and 7 8
-1 and - 7 - 8

We found the factors whose product is 7 and whose sum is - 8. x^2- 8x+7 ⇔ (x-1)(x-7)

b To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.
y^2-2y- 15 In this case, we have - 15. This is a negative number, so for the product of the constant terms in the factors to be negative, these constants must have the opposite sign — one positive and one negative.
Factor Constants Product of Constants
1 and - 15 - 15
-1 and 15 - 15
3 and - 5 - 15
-3 and 5 - 15

Next, let's consider the coefficient of the linear term. y^2- 2y- 15 For this term, we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, - 2.

Factors Sum of Factors
1 and - 15 - 14
-1 and 15 14
3 and - 5 - 2
-3 and 5 2

We found the factors whose product is - 15 and whose sum is - 2. y^2- 2y- 15 ⇔ (y+3)(y-5)

c We want to factor the given equation. Let's start factoring by first identifying the greatest common factor (GCF).

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. In this case, the GCF is 7.
7x^2-63
7* x^2- 7* 9
7(x^2-9)

The result of factoring out a GCF from the given expression is a binomial with a leading coefficient of 1. 7( x^2-9) Let's temporarily only focus on this binomial, and we will bring back the GCF after factoring.

Factor the Expression

Do you notice that the binomial is a difference of two perfect squares? This can be factored using the difference of squares method. a^2 - b^2 ⇔ (a+b)(a-b) To do so, we first need to express each term as a perfect square.

Expression x^2-9
Rewrite as Perfect Squares x^2 - 3^2
Apply the Formula (x+3)(x-3)

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. 7(x+3)(x-3)

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+10x+8 ⇔ 3x^2+10x+8 We have that a= 3, b=10, and c=8. 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=8, the value of a c is 3* 8=24.
  2. Find factors of a c. Since a c=24, 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=10, 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 &24 &1+24 &25 2 &12 &2+12 &14 3 &8 &3+8 &11 4 & 6 & 4 + 6 &10

  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+10x+8 ⇕ 3x^2+ 4x+ 6x+8
Finally, we will factor the last expression obtained.
3x^2+4x+6x+8
x(3x+4)+6x+8
x(3x+4)+2(3x+4)
(x+2)(3x+4)