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Quiz 1 and 2 Solutions:
Quiz #1 (each problem is worth 2 points)
1) GCF (36; 63) =
Notice that 36 = 22 32 and 63 = 32 7:
The only prime number the have in common is 3 ;
and the lowest power of the 3 is 2: Therefore, GCF (36; 63) = 9:
2) Write y 2 (x 2) + (x 2) as a product
Since each term shares a common factor of (x 2) we can combine
the coe¢ cients. We have y 2 (x 2) + (x 2) = (y 2 + 1)(x 2)
3) Factor each of the following:
a) x2 + 5x + 6 = (x + 2)(x + 3) since 2 3 = 6 and 2 + 3 = 5:
b) x2 + 6x + 9 = (x + 3)(x + 3) = (x + 3)2 since 3 3 = 9 and 3 + 3 = 6:
4) GCF (9x2 y; 3xy) =
Notice that GCF (9; 3) = 3; GCF (x2 ; x) = x; and GCF (y; y) = y:
So, the GCF (9x2 y; 3xy) is their product, i.e. GCF (9x2 y; 3xy) = 3 x y = 3xy:
Quiz #2 (each problem is worth 4 points)
1) Factor 2x2 + 7x + 3 by grouping
First we take the product the outsides, i.e. 2 3 = 6:
Now we look for two numbers whose product is 6 but whose sum is the middle
term, i.e. 7: Since 6 = 6 1 and 6 + 1 = 7 these are our magic numbers.
Therefore,
2x2 + 7x + 3 = 2x2 + 6x + x + 3
= 2x(x + 3) + 1 (x + 3)
= (2x + 1)(x + 3)
Notice that the order of 6x and x doesn’t matter. We also could have had:
2x2 + 7x + 3 = 2x2 + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (x + 3)(2x + 1)
2) Factor 8x2 + 6x 27 any way you can
By trial and error we get (4x + 9)(2x 3). Notice that 4x 2x = 8x2 , 9 ( 3) = 27;
and by the outside inside trick, 4x( 3) + 9(2x) = 12x + 18x = 6x as desired.
Alternatively, we could have factored by grouping, but with such big numbers it would
not be very e¢ cient.