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Think of the process of factoring as multiplying two binomials in reverse.
Is there a greatest common factor between all of the terms in the given expression? If so, you should factor that out first.
Is there a greatest common factor between all of the terms in the given expression? If so, you should factor that out first.
x=6 or x=7
x=2/3 or x=- 4
x=0 or x=5
x=3 or x=- 5
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.
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
Since the equation is already written in factored form, we can now use the Zero Product Property.
Use the Zero Product Property
(I): LHS+6=RHS+6
(II): LHS+7=RHS+7
We found that x=6 or x=7.
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= 6
Calculate power
Multiply
Add and subtract terms
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= 7
Calculate power
Multiply
Add and subtract terms
Again, we created a true statement. x=7 is indeed a solution of the equation.
We want to solve the given equation for x. To do this, we can use factoring. Then, we will use the Zero Product Property.
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)
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
Finally, we will factor the last equation obtained.
Factor out x
Factor out 4
Factor out (3x-2)
Now, the equation is written in a factored form.
Since the equation is already written in factored form, we can now use the Zero Product Property.
Use the Zero Product Property
(II): LHS-4=RHS-4
We found that x= 23 or x=- 4.
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.
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.
Since the equation is already written in factored form, we can now use the Zero Product Property.
Use the Zero Product Property
(I): .LHS /2.=.RHS /2.
(II): LHS+5=RHS+5
We found that x=0 or x=5.
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.
x= 0
Calculate power
Zero Property of Multiplication
Subtract term
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.
Again, we created a true statement. x=5 is indeed a solution of the equation.
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).
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.
Split into factors
Factor out 4
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
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
Since the equation is already written in factored form, we can now use the Zero Product Property.
.LHS /4.=.RHS /4.
Use the Zero Product Property
(I): LHS+3=RHS+3
(II): LHS-5=RHS-5
We found that x=3 or x=- 5.
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.
x= 3
Calculate power
Multiply
Add and subtract terms
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.
x= - 5
(- a)^2 = a^2
a(- b)=- a * b
Multiply
Subtract terms
Again, we created a true statement. x=- 5 is indeed a solution of the equation.