Of angle JEK is complementary to angle DFH, then
1 answer:
You might be interested in
Answer:
2) x³ (x² + 1) (x + 1)
3) (x - 6) (x + 6)
6) (x + 3) (x - 4)
8) (5x + 3) (x - 2)
Step-by-step explanation:
I'm providing the solutions for numbers 2, 3, 6, and 8, in order to give you the opportunity to apply and practice the techniques demonstrated here for numbers 9 and 10.
<h3>2) x⁶ + x⁵ + x⁴ + x³ = 0 </h3>
Factor out the highest possible common term, applying the product rule of exponents,
x³ (x³ + x² + x¹ + 1) = 0
Next, apply the special factoring technique into the exponential expressions inside the parenthesis), the factor by grouping:
The factors for the given polynomial are: x³ (x² + 1) (x + 1)
<h3> 3) x² - 36 = 0 </h3>
Use the <u>difference of two squares factoring</u>, where it states: u² - v² = (u - v) (u + v)
Since 36 is a perfect square of 6², then we can establish the following factors using the difference of squares:
Factors: (x - 6) (x + 6).
<h3>6) x² - x - 12 = 0 </h3>
For this quadratic trinomial, find factors that will produce a product of <em>a</em> × <em>c</em> (1 × -12) and sum of b (-1):
The possible factors of -12 that also results to a sum of -1 are:
3 × -4 = -1
Hence, the possible factors of x² - x - 12 = 0 could be (x + 3) (x - 4). Verify whether this is correct by performing the FOIL method:
(x + 3) (x - 4) = x² - 4x + 3x - 12 = x² - x - 12
Therefore, the factors of x² - x - 12 = 0 are: (x + 3) (x - 4).
<h3>8) 5x² - 7x - 6 = 0 </h3>
where a = 5, b = -7, and c = -6
Given this quadratic trinomial, we can use the factoring by grouping, where it states: find the values of u and v such that the product of uv is the same as the product of <em>ac</em>: u × v = a × c, and that the sum of (<em>u</em> + <em>v </em>) is the same as <em>b</em>.
In other words, we must find the factors of u × v that have a product of a × c and a sum of b:
<em>u × v </em>= <em>a × c</em> = 5 × (-6) = -30
u + v = b = -7
The possible factors of -30 are:
-3 × 10 = -30
3 × -10 = -30
Out of these factors, the factors 3 × -10 = -30 will have the same sum as the value of b = -7:
3 + (-10) = -7
Hence, u = 3, v = -10.
Now, we can group the trinomial as follows:
(<em>ax</em>² + <em>ux </em>) + (<em>vx </em>+ <em>c </em>)
(5x² + 3x) + (-10x - 6)
For the last step, factor out the common terms from both groups:
(5x² + 3x) + (-10x - 6)
x (5x + 3) - 2(5x + 3)
Combine common terms together: (5x + 3) and (x - 2)
Therefore, the factors of the given quadratic trinomial are: (5x + 3) (x - 2).