Mth 271
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Calculus: An Applied Approach 7th edition. Pg. 0-15 section 0.3 example 5. Simplify each expression by factoring.
a.) 3(x+1)^1/2 (2x - 3)^5/2 + 10(x+1)^3/2 (2x-3)^3/2.
The next steps I do not understand how they did. Next step: =
(x+1)^1/2(2x-3)^3/2 [3(2x-3) + 10(x+1)]
It might be easier to see how this was done if you multiply the last expression out.
(x+1)^1/2(2x-3)^3/2 [3(2x-3) + 10(x+1)] =
(x+1)^1/2(2x-3)^3/2 * 3(2x-3) + (x+1)^1/2(2x-3)^3/2 * 10(x+1) by the distributive law.
(x+1)^1/2(2x-3)^3/2 * 3(2x-3) = 3(x+1)^1/2 (2x - 3)^5/2 and
(x+1)^1/2(2x-3)^3/2 * 10(x+1) = 10(x+1)^3/2 (2x-3)^3/2.
So
(x+1)^1/2(2x-3)^3/2 [3(2x-3) + 10(x+1)] = 3(x+1)^1/2 (2x - 3)^5/2 + 10(x+1)^3/2 (2x-3)^3/2
Now, to factor 3(x+1)^1/2 (2x - 3)^5/2 + 10(x+1)^3/2 (2x-3)^3/2 you use the same ideas, but in reverse order.
We first note than both terms contain powers of x+1 and of 2x - 3.
Whichever power of x + 1 is the lesser, that power will be common to both terms of our expression. In this case 1/2 is the common power, so (x + 1)^(1/2) is common to both terms.
Whichever power of 2x - 3 is the lesser, that power will be common to both terms of our expression. In this case 3/2 is the common power, so (2x - 3)^(3/2) is common to both terms.
So we factor (x + 1)^(1/2) * (2x - 3)^(3/2) out of both terms of the original expression. (To figure out what each remaining term will be, just divide each term by (x + 1)^(1/2) * (2x - 3)^(3/2)).
How did they move the 3 to the last part of the equation and get ^3/2 in place of the ^5/2?
Next step:=
(x+1)^1/2 (2x-3)^3/2 (6x-9 +10x+10)
This step I get just not how they got here from the last step. Final step: =
(x+1)^1/2 (2x+3)^3/2 (16x+1)
6x-9 +10x+10 = 6x + 10 x - 9 + 10 = 16 x + 1.
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Very well-expressed questions. Hopefully my notes will help. Let me know if you need further clarification.