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Angle 1 = 30°
Angle 2 = 90°
⇒ Angle 3 = 60°
So, it's right triangle. We can set the length of one side and get all other sides.
So, we have 1 triangle. If there are 2 or more triangles with the same data, all the triangles will be congruent because of : <span>Two triangles are </span><span>congruent if </span>"<span>ASA: Two interior angles and the included side in a triangle have the same measure and length, respectively, as those in the other triangle</span><span>.</span>"
11. Factoring and solving equations
- A. Factor-
1. Factor 3x2 + 6x if possible.
Look for monomial (single-term) factors first; 3 is a factor of both 3x2
and 6x and so is x . Factor them out to get
3x2 + 6x = 3(x2 + 2x1 = 3x(x+ 2) .
2. Factor x2 + x - 6 if possible.
Here we have no common monomial factors. To get the x2 term
we'll have the form (x +-)(x +-) . Since
(x+A)(x+B) = x2 + (A+B)x + AB ,
we need two numbers A and B whose sum is 1 and whose product is
-6 . Integer possibilities that will give a product of -6 are
-6 and 1, 6 and -1, -3 and 2, 3 and -2.
The only pair whose sum is 1 is (3 and -2) , so the factorization is
x2 + x - 6 = (x+3)(x-2) .
3. Factor 4x2 - 3x - 10 if possible.
Because of the 4x2 term the factored form wli be either
(4x+A)(x +B) or (2x+A)(2x+B) . Because of the -10 the integer possibilities
for the pair A, B are
10 and -1 , -10 and 1 , 5 and -2 . -5 and 2 , plus each of
these in reversed order.
Check the various possibilities by trial and error. It may help to write
out the expansions
(4x + A)(x+ B) = 4x2 + (4B+A)x + A8
1 trying to get -3 here
(2x+A)(2x+B) = 4x2 + (2B+ 2A)x + AB
Trial and error gives the factorization 4x2 - 3x - 10 - (4x+5)(x- 2) .
4. Difference of two squares. Since (A + B)(A - B) = - B~ , any
expression of the form A' - B' can be factored. Note that A and B
might be anything at all.
Examples: 9x2 - 16 = (3x1' - 4' = (3x +4)(3x - 4)
x2 - 29 = x2 - (my)* = (x+ JTy)(x- my)
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