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| 11 Theory slides |
| 9 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Consider the following quadratic equation for x. 2 x^2+2 x-4=0
Can this equation be solved by taking square roots? If not, how can it be solved algebraically — without graphing?By the Zero Product Property, at least one of the factors must equal 0. Therefore, each of the factors can be set equal to 0. 3x-9=0 or x+5=0 Note that the word connecting the equations is or. By setting the factors equal to 0, new equations are created. There are as many equations as there were factors — in this case there are two.
The following applet shows a quadratic equation whose left-hand side is written in factored form. Using the Zero Product Property find the solutions of the equation. Round to the nearest tenth if necessary.
After factoring out the GCF, the left-hand side of the equation is in factored form. Now, the steps of solving an equation using the Zero Product Property will be followed. By the Zero Product Property each of the factors can be set equal to 0. 3x(x-4)=0 ⇒ lc3x=0 & (I) x-4=0 & (II)
Dolphins jump out of the water to improve their navigation and to see the surface of the ocean. They also do it for fun.
The dolphin is in the water when h=0.
Rewrite - 5t^2 as t(- 5t)
Commutative Property of Multiplication
Factor out t
seeits factors. Consider the following expression. 8x^2+ 26x+ 6 Here, a= 8, b= 26, and c= 6. There are six steps to factor this trinomial.
It is known that a= 4 and c= 3, so a c=12>0. Therefore, the factors must have the same sign. Also, b= 13. Since the sum of the factors is positive and they must have the same sign, both factors must be positive. All positive factor pairs of 12 can now be listed and their sums checked.
Factors of a c | Sum of Factors |
---|---|
1 and 12 | 1+12=13 ✓ |
2 and 6 | 2+6=8 * |
3 and 4 | 3+4=7 * |
In this case, the correct factor pair is 1 and 12. The following table sums up how to determine the signs of the factors based on the values of ac and b.
ac | b | Factors |
---|---|---|
Positive | Positive | Both positive |
Positive | Negative | Both negative |
Negative | Positive | One positive and one negative. The absolute value of the positive factor is greater. |
Negative | Negative | One positive and one negative. The absolute value of the negative factor is greater. |
Such analysis makes the list of possible factor pairs shorter.
The above method can be used to factor any quadratic trinomial. It is particularly useful for trinomials in the forms ax^2+bx or ax^2+c. However, in these two cases there are other approaches that might be used as well.
Please note that not every quadratic trinomial can be factored. When a pair of integers whose product is equal to a* c and whose sum is b cannot be found, the trinomial cannot be factored using the described method.
x^2-x+1, a* c=1, b=- 1 | |
---|---|
Product of Factors | Sum of Factors |
1* 1=1 | 1+1=2 * |
- 1(- 1)=1 | - 1+(- 1)=- 2 * |
Think of a quadratic equation in standard form. ax^2+bx+c=0
If its left-hand side is factorable, the equation can be solved using the Zero Product Property.Zosia is a student studying how many Dolphins swim past a certain area of cove. She marks off a rectangle using rope where she will count how many dolphins pass through in any given day. The study area's rectangle is equal to 42 square meters. However, the dimensions of the rectangle are unknown.
Based on the given information, write and solve a quadratic equation for x.
Distribute (4x-7)
Distribute 2x & - 1
Subtract term
LHS-42=RHS-42
Pair of Factors | Sum of Factors |
---|---|
1 and - 280 | 1+(- 280)=- 279 * |
2 and - 140 | 2+(- 140)=- 138 * |
4 and - 70 | 4+(- 70)=- 66 * |
5 and - 56 | 5+(- 56)=- 51 * |
7 and - 40 | 7+(- 40)=- 33 * |
8 and - 35 | 8+(- 35)=- 27 * |
10 and - 28 | 10+(- 28)=- 18 ✓ |
14 and - 20 | 14+(- 20)=- 6 * |
Split into factors
Factor out 2x
Factor out - 7
Use the Zero Product Property
(I): LHS-5=RHS-5
(I): .LHS /4.=.RHS /4.
(I): Put minus sign in front of fraction
(I): Calculate quotient
2x-1 | 4x-7 | |
---|---|---|
x=- 1.25 | 2( - 1.25)-1=- 3.5 * | 4( - 1.25)-7=- 12 * |
x=3.5 | 2( 3.5)-1=6 ✓ | 4( 3.5)-7=7 ✓ |
It can be seen that x=- 1.25 results in a negative width and length, which is not logical for the dimensions of a rectangle. Therefore, the only valid solution is x=3.5. The width of the rectangle is 6 meters and the length is 7 meters.
Rewrite the equation in standard form. Factor the left-hand side of the rewritten equation.
Pair of Factors | Sum of Factors |
---|---|
- 1 and - 225 | - 1+(- 225)=- 226 * |
- 3 and - 75 | - 3+(- 75)=- 78 * |
- 5 and - 45 | - 5+(- 45)=- 50 * |
- 9 and - 25 | - 9+(- 25)=- 34 * |
- 15 and - 15 | - 15+(- 15)=- 30 ✓ |
Rewrite - 30x as - 15x-15x
Split into factors
Rewrite 25 as - 5(- 5)
Factor out - 5
Factor out (3x-5)
Zosia, waiting for dolphins to swim by, decides to throw a pebble from an 18-foot cliff into the sea. The following quadratic function describes the height of the pebble y above water x seconds after the it was thrown. y=- 18x^2+27x+18 The height y is expressed in feet.
Split into prime factors
Commutative Property of Multiplication
Multiply
Factor out 9
.LHS /9.=.RHS /9.
Pair of Factors | Sum of Factors |
---|---|
- 1 and 4 | - 1+4=3 ✓ |
- 2 and 2 | - 2+2=0 * |
Write as a sum
Factor out (2x+1)
Use the Zero Product Property
LHS-30=RHS-30
Rearrange equation
Pair of Factors | Sum of Factors |
---|---|
1 and 24 | 1+24=25 * |
2 and 12 | 2+12=14 * |
3 and 8 | 3+8=11 * |
4 and 6 | 4+6=10 * |
As seen above none of the pairs adds up to 9. Therefore, the equation cannot be factored using integers. This might mean that the equation has no solutions. To confirm, the function corresponding to the equation will be graphed. Equation: & - 6x^2+9x-4=0 Function: & y=- 6x^2+9x-4 The quadratic function in standard form will be now graphed. Recall the definitions of the axis of symmetry, the vertex, and the y-intercept.
This lesson showed how to use factoring and the Zero Product Property to solve quadratic equations. The factoring method can be also used to solve other types of equations, particularly cubic equations of the following form. ax^3+bx^2+cx=0 Since the constant term d is equal to 0, x can be factored out in the equation. ax^3+bx^2+cx=0 ⇕ x(ax^2+bc+c)=0 Next, two equations are obtained by the Zero Product Property. x(ax^2+bc+c)=0 ⇓ lcx=0 & (I) ax^2+bx+c=0 & (II)
Equation (I) is already solved. Equation (II) is a quadratic equation that can be solved by factoring or by using some other known method. In general, factoring allows the splitting of a more complicated equation into simpler ones.
We are told that both fractions describe the same number. This means that we can set them equal to each other. 18+b/2b=32-b/3b This expression is an equation which can be solved for b. Let's do it!
The equation can now be solved using the Zero Product Property. Let's do it!
This work shows that b can be equal to 0 or 2. However, if we take b=0 the denominator becomes 0, so the fraction would be undefined. This means that the only acceptable solution is b=2. Since the expressions are equivalent, we can substitute 2 into either expression.
We have found that the fractions are equal to 5.
Consider the following diagram.
Let's begin by recalling the formula for the area of a rectangle. A=l w We can see from the diagram that the length l of the rectangle is 4x-2 and its width w is 2x-3. Let's substitute these values into the formula.
We are told that the area of the rectangle is 6. Let's substitute this into our equation. A=(4x-2)(2x-3) ⇓ 6=(4x-2)(2x-3) We will now solve this equation for x. First, we will rewrite it so that the left-hand side is in factored form and the right-hand side is equal to 0.
We can now use the Zero Product Property to solve this equation. Let's do it!
The solutions to the equation are x=0 and x=2. However, if we use x=0 we would have negative lengths, which makes no sense, so we discard such solution. Therefore, the answer is x=2.
Having h(t) equal to 0 means that the soccer ball is on the ground at that time. Knowing this, we can set h(t) equal to 0 to write an equation to solve for t. Let's write the equation, and then solve it!
The solutions to the equation are t=0 and t=b. Note that the solution t=0 is the instant of the kick, so the soccer ball is still on the ground, which makes perfect sense. h(0)=0 However, we are asked to find how long it takes the ball to reach the ground after being kicked, so the answer is t=b seconds.