Algebra 2 by Richard Wright
Are you not my student and
has this helped you?
Objectives:
SDA NAD Content Standards (2018): AII.5.1, AII.5.2, AII.6.3, AII.6.4
Quadratic equations are a type of polynomial equations. The simplest way to solve quadratic equations was by factoring, if they were factorable. Likewise, the simplest way to the solve polynomial equations is by factoring.
Factoring is unmultiplying, taking the product and separating it into the parts that were multiplied.
To factor a polynomial,
Factor 2x2 − 50.
Solution
Start by looking for the common factor. In this case the greatest common factor is 2.
2(x2 − 25)
The remaining factor has two terms, so use a special formula. It is squared, so it is a difference of squares with a = x and b = 5.
a2 − b2 = (a − b)(a + b)
2(x − 5)(x + 5)
Factor 8x3 + 343.
Solution
Start by looking for a common factor. There is none.
There are two terms, so use a special formula. The x is cubed, so check to see if the other numbers are also cubes.
8x3 + 343
(2x)3 + 73
This is a sum of cubes with a = 2x and b = 7.
a3 + b3 = (a + b)(a2 − ab + b2).
(2x + 7)((2x)2 − (2x)(7) + (7)2)
(2x + 7)(4x2 − 14x + 49)
The second factor is still squared, so try factoring it. However, it is not factorable.
Factor 3x3 − 6x2 − 24x.
Solution
Start by looking for a common factor, in this case each term is divisible by 3x.
3x3 − 6x2 − 24x
3x(x2 − 2x − 8)
The remaining factor has three terms, so use guess-and-check.
What two terms make the first term: x2? x·x. These make the firsts.
3x(x )(x )
What two terms make the last term: −8? −4·2. These make the lasts.
3x(x − 4)(x + 2)
Check the to see if the inner + outer = middle.
2x + (−4x) = −2x
This works. Thus, the factorization is
3x(x − 4)(x + 2)
Factor x3 + 2x2 − 16x − 32.
Solution
Start by looking for a common factor. There are none.
There are four terms, so factor by grouping.
Group the first two and last two terms. Leave the minus sign inside the second group.
x3 + 2x2 − 16x − 32
(x3 + 2x2) + (−16x − 32)
Factor the common factor out of each group.
x2(x + 2) − 16(x + 2)
The common factor is now (x + 2). Factor it out.
(x + 2)(x2 − 16)
The x2 – 16 is a difference of squares.
(x + 2)(x − 4)(x + 4)
Solving by factoring relies on the zero-product property which says if ab = 0, then either a = 0 or b = 0. Therefore, the polynomial equation is setup so that it equals zero. Then, if it can be written as a product of factors, the solutions will be the values of x that make the factors equal zero.
To solve a polynomial by factoring,
Solve 3x4 − 24x = 0.
Solution
Start by looking for a common factor. In this case 3x.
3x4 − 24x = 0
3x(x3 − 8) = 0
The x3 – 8 is a difference of cubes since 8 is 23 with a = x and b = 2.
a3 − b3 = (a − b)(a2 + ab + b2)
3x(x − 2)(x2 + 2x + 4) = 0
The x2 + 2x + 4 is not factorable, so the factoring is done.
Set each factor equal to zero.
3x = 0
x = 0
Or
x − 2 = 0
x = 2
Or
x2 + 2x + 4 = 0
Use the quadratic formula because this is not factorable.
$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$
$$ x=\frac{-2\pm\sqrt{2^2-4\left(1\right)\left(4\right)}}{2\left(1\right)} $$
$$ x=\frac{-2\pm\sqrt{-12}}{2} $$
$$ x=\frac{-2\pm2\sqrt3i}{2} $$
x = \(-\mathbf{1}\pm\sqrt{\mathbf{3}}\)i
The solutions are 0, 2, and \(-\mathbf{1}\pm\sqrt{\mathbf{3}}\)i.
Solve 5x3 − 15x2 + 7x − 21 = 0.
Solution
Start by looking for a common factor. There is none.
There are four term, so factor by grouping. Group the first two and last two terms.
5x3 − 15x2 + 7x − 21 = 0
(5x3 − 15x2) + (7x − 21) = 0
Factor the common factor out of each group.
5x2 (x − 3) + 7(x − 3) = 0
The x – 3 is now the common factor, so factor it out.
(x − 3)(5x2 + 7) = 0
The 5x2 + 7 is not factorable, so the factoring is done.
Set each factor equal to zero and solve.
x − 3 = 0
x = 3
Or
5x2 + 7 = 0
5x2 = −7
x2 = \(-\frac{7}{5}\)
$$ x=\pm\sqrt{-\frac{7}{5}} $$
x = \(±\frac{\sqrt{\mathbf{35}}}{\mathbf{5}}\)i
The solutions are 3 and \(±\frac{\sqrt{\mathbf{35}}}{\mathbf{5}}\)i.
180 #1, 5, 9, 13, 17, 21, 25, 29, 33, 49, 53, 188 #1, 3, 5, 7, Mixed Review = 20
Mixed Review
, x-ints: (−3, 0), (1/2, 0), (5, 0)