Tuesday, March 31, 2020

Algebra 2 Problems of the Day (Jan 2020)

Daily Algebra 2 questions and answers.

More Algebra 2 problems.

January 2020, Part II

All Questions in Part II are worth 2 credits. Work must be shown. Partial credit is given.


25. For n and p > 0, is the expression (p2n1/2)8SQRT(p5n4) equivalent to p18n6SQRT(p)? Justify your answer.

Answer:
Multiply the exponents in the parentheses of the first part of the first expression by 8. Multiply the exponents in the second part by 1/2. Then add the exponents of each matching variable.
(p16n4)(p5/2n2)
= p18 1/2n6
= p18n6SQRT(p)

Yes, they are equivalent.

Note: You must include the concluding statement to get full credit. However, the concluding statement, by itself, with no work, is not worth any points.





26. Show why x - 3 is a factor of m(x) = x3 - x2 - 5x - 3. Justify your answer.

Answer:
There are several methods.
You can show that x - 3 is a factor m(x) by calculating m(3), which will give you 0 if it is a factor.
m(3) = (3)3 - (3)2 - 5(3) - 3 = 27 - 9 - 15 - 3 = 0
Since 3 is a zero of the function, x - 3 is a factor.

You can also show this by factoring or dividing the function. You can use, for example, long division or the grid method.

Look at the image below.
Put x and -3 on the left and x3 in the first box.
Divide x into x3 and you get x2. Write that on top. Now multiply x2 by -3, which has a product of -3x3. Write that in the bottom of the first column.

We have -3x2 but the original expression has -x2, so we have to add 2x2 in the top box of the second column. Repeat the division and multiplication.

When you get to the last box, you have -3, which is what we need. There is no remainder.

(x - 3) is a factor of m(x) because it divides evenly with no remainder.





Comments and questions welcome.

More Algebra 2 problems.

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