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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.
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.
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 -5t2 as t(-5t)
Commutative Property of Multiplication
Factor out t
seeits factors. Consider the following expression.
It is known that a=4 and c=3, so ac=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 ac | 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 ax2+bx or ax2+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.
x2−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 × |
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)
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
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 × |
Since we know that x=5 is a solution to the given equation we can substitute it to find b. Let's do it!
We found that b=7. Let's include it and rewrite the equation. x^2-7x+10=0 We are one step closer. Now, let's have r represent the unknown root. By using the Zero Product Property we can write an equivalent equation. (x-r)(x-5)=0 We can expand this equation so it looks like the given equation written in standard form of a quadratic equation.
Since the equations are equivalent, the coefficients have to be equal. x^2- 7&x+10=0 x^2-(5+r)&x+5r=0 This means that 5+r=7 and 10=5r. Both equations can then be solved for r. c|c 5+r=7 & 5r=10 ⇕ & ⇕ r=2 & r=2 They both result in r=2, so the second root is x=2.
Let's begin by noticing that x can be factored out from the left-hand side of the equation. x^3-5x^2+6x=0 ⇕ x(x^2-5x+6)=0 By the Zero Product Property x=0 is one of the roots of the equation. The other solutions can be found by setting x^2-5x+6 equal to 0 and solving the resulting equation. Let's do it!
We determined that the other two roots are x=2 and x=3.