Core Connections Integrated II, 2015
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2. Section 5.2
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Exercise 100 Page 296

Practice makes perfect
a We want to solve the given equation for x. To do this we can start by using factoring. Then we will use the Zero Product Property.

Factoring

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^2-13x+42=0 In this case we have 42. 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 42 42
-1 and -42 42
2 and 21 42
-2 and -21 42
3 and 14 42
-3 and -14 42
6 and 7 42
-6 and -7 42

Next, let's consider the coefficient of the linear term. x^2-13x+42=0 For this term, we need the sum of the factors that produced the constant term equal the coefficient of the linear term, - 13.

Factors Sum of Factors
1 and 42 43
-1 and -42 - 43
2 and 21 23
-2 and -21 - 23
3 and 14 17
-3 and -14 - 17
6 and 7 13
-6 and -7 - 13

We found the factors whose product is 42 and whose sum is - 13. x^2-13x+42=0 ⇕ (x-6)(x-7)=0

Zero Product Property

Since the equation is already written in factored form, we can now use the Zero Product Property.
(x-6)(x-7)=0
lcx-6=0 & (I) x-7=0 & (II)
lx=6 x-7=0
lx_1=6 x_2=7
We found that x=6 or x=7.

Checking Our Answer

We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=6.
x^2-13x+42=0
6^2-13( 6)+42? =0
â–Ľ
Simplify
36-13(6)+42? = 0
36-78+42? = 0
0=0 âś“
Substituting and simplifying created a true statement, so we know that x=6 is a solution of the equation. Let's move on to x=7.
x^2-13x+42=0
7^2-13( 7)+42? = 0
â–Ľ
Simplify
49-13(7)+42? = 0
49-91+42? = 0
0=0 âś“
Again, we created a true statement. x=7 is indeed a solution of the equation.
b We want to solve the given equation for x. To do this, we can use factoring. Then, we will use the Zero Product Property.

Factoring

Here we have a quadratic equation of the form 0=ax^2+bx+c, where |a| ≠ 1 and there are no common factors. To factor this equation 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. 0=3x^2+10x-8 ⇕ 0= 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 equation.

  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 ac=- 24, which is negative, factors of a c need to have opposite signs — one positive and one negative. Since b=10, 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 & 24 &-1 + 24 &23 - 2 & 12 & - 2 + 12 &10 - 3 &8 &-3 + 8 &5 - 4 &6 &-4 + 6 &2

  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. 0=3x^2+10x-8 ⇕ 0=3x^2 - 2x + 12x-8
Finally, we will factor the last equation obtained.
0=3x^2-2x+12x-8
0=x(3x-2)+12x-8
0=x(3x-2)+4(3x-2)
0=(3x-2)(x+4)
Now, the equation is written in a factored form.

Zero Product Property

Since the equation is already written in factored form, we can now use the Zero Product Property.
0=(3x-2)(x+4)
lc3x-2=0 & (I) x+4=0 & (II)
â–Ľ
(I): Solve for x
l3x=2 x+4=0
lx= 23 x+4=0
lx_1= 23 x_2=- 4
We found that x= 23 or x=- 4.
c We want to solve the given equation for x. To do this, we can use factoring. We will start from identifying the greatest common factor (GCF). Then, we will use the Zero Product Property.

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 2x.
2x^2-10x=0
2x* x- 2x* 5
2x(x-5)=0

Now, the equation is written in a factored form.

Zero Product Property

Since the equation is already written in factored form, we can now use the Zero Product Property.
2x(x-5)=0
lc2x=0 & (I) x-5=0 & (II)
lx=0 x-5=0
lx_1=0 x_2=5
We found that x=0 or x=5.

Checking Our Answer

We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=0.
2x^2-10x=0
2( 0)^2-10( 0)? =0
â–Ľ
Simplify
2(0)-10(0)? = 0
0-0? = 0
0=0 âś“
Substituting and simplifying created a true statement, so we know that x=0 is a solution of the equation. Let's move on to x=5.
2x^2-10x=0
2( 5)^2-10( 5)? =0
â–Ľ
Simplify
2(25)-10(5)? = 0
50-50? = 0
0=0 âś“
Again, we created a true statement. x=5 is indeed a solution of the equation.
d We want to solve the given equation for x. To do this, we can start by using factoring. Then, we will use the Zero Product Property. Let's start factoring by identifying the greatest common factor (GCF).

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 4.
4x^2+8x-60=0
4* x^2+ 4* 2x- 4* 15=0
4(x^2+2x-15)=0
The result of factoring out a GCF from the given expression is a trinomial with a leading coefficient of 1. 4( x^2+2x-15)=0

Let's temporarily only focus on this trinomial, and we will bring back the GCF after factoring.

Factor the Expression

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+2x- 15=0 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. x^2+2x- 15=0 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. x^2+2x- 15=0 ⇕ (x-3)(x+5)=0 Wait! Before we finish, remember that we factored out a GCF from the original equation. To fully complete the factored equation, let's reintroduce that GCF now. 4(x-3)(x+5)=0

Zero Product Property

Since the equation is already written in factored form, we can now use the Zero Product Property.
4(x-3)(x+5)=0
(x-3)(x+5)=0
lcx-3=0 & (I) x+5=0 & (II)
lx=3 x+5=0
lx_1=3 x_2=- 5
We found that x=3 or x=- 5.

Checking Our Answer

We can substitute our solutions back into the given equation and simplify to check if our answers are correct. We will start with x=3.
4x^2+8x-60=0
4( 3)^2+8( 3)-60? =0
â–Ľ
Simplify
4(9)+8(3)-60? = 0
36+24-60? = 0
0=0 âś“
Substituting and simplifying created a true statement, so we know that x=3 is a solution of the equation. Let's move on to x=- 5.
4x^2+8x-60=0
4( - 5)^2+8( - 5)-60? =0
â–Ľ
Simplify
4(25)+8(- 5)-60? = 0
4(25)-40-60? = 0
100-40-60? = 0
0=0 âś“
Again, we created a true statement. x=- 5 is indeed a solution of the equation.