Targeted Review Solutions

Name the vertices formed from the system of linear inequalities in the given graph.

$\{(0, 7), (8, 2), (5, 0), (0, 4)\}$

Using the objective function $f (x, y) = 2 x + y$ , determine the minimum and maximum using the given graph.

$f (0, 7) = 2 (0) + (7) f (0, 7) = 7 f (8, 2) = 2 (8) + (2) f (8, 2) = 18 maximum f (5, 0) = 2 (5) + (0) f (5, 0) = 10 f (0, 4) = 2 (0) + (4) f (0, 4) = 4 minimum$

Write the polynomial identity using the variables x and y.

Difference of two squares

$x^{2} - y^{2} = (x + y) (x - y)$

Difference of cubes

$x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$

Divide $15 x^{5} + 9 x^{3} + 2 x^{2} - 6 by x$

$\frac{15 x^{5} + 9 x^{3} + 2 x^{2} - 6}{6 x} \frac{15 x^{5}}{6 x} + \frac{9 x^{3}}{6 x} + \frac{2 x^{2}}{6 x} - \frac{6}{6 x}$

$\frac{5 x^{4}}{2} + \frac{3 x^{2}}{2} + \frac{x}{3} - \frac{1}{x}$

Determine the value that would be used for synthetic substitution or synthetic division from the given binomials.

1. 1. $a - 5$
  2. $3 b + 1$
  3. $3 - c$

$a - 5 = 0 a = 5$
$3 b + 1 = 0 3 b = - 1 b = - \frac{1}{3}$
$3 - c = 0 c = 3$

Simplify. Name any restrictions on the denominator.

$\frac{15 x^{4} - 15 y^{4}}{5 x^{2} - 5 y^{2}}$

$\frac{15 (x^{4} - y^{4})}{5 (x^{2} - y^{2})} \frac{15 (x^{2} - y^{2}) (x^{2} + y^{2})}{5 (x + y) (x - y)} x \neq \pm y \frac{15 (x + y) (x - y) (x^{2} + y^{2})}{(x + y) (x - y)} \frac{{}^{3}15 (x + y) (x - y) (x^{2} + y^{2})}{{}^{1}5 (x + y) (x - y)}$

$3 (x^{2} + y^{2}) x \neq \pm y$

Note

Solve for x. The numerator and denominator are the difference of two squares.

$\frac{8 x^{3} + 12 x}{12 x^{3} + 34 x^{2} + 24 x}$

$\frac{4 x (2 x + 3)}{2 x (6 x^{2} + 17 x + 12)} \frac{4 x (2 x + 3)}{2 x (2 x + 3) (3 x + 4)} x \neq - \frac{3}{2}, - \frac{4}{3}, 0 \frac{{}^{2}{4 x} (2 x + 3)}{2 x (2 x + 3) (3 x + 4)}$

$\frac{2}{3 x + 4}, x \neq - \frac{3}{2}, - \frac{4}{3}, 0$

Multiple Choice

Select all that apply.
When a region is unbounded for linear programming this means:

the region will not be completely enclosed.
the region does not exist.
the region will continue infinitely in at least one direction.
nothing is known about the problem.

Note

The unchecked options do not correctly describe an unbounded region.

Determine the restrictions on the denominator for a triangle with an area of:
$\frac{25 - 4 x^{2}}{3 x^{2} - 13 x - 10}$

$- 5, \frac{2}{3}$
$- 2, \frac{5}{3}$
$- \frac{2}{3}, 5$
$\pm \frac{5}{2}$

$(3 x + 2) (x - 5) = 0 x = - \frac{2}{3}, 5$

Note

This option is the answer if the middle term in the denominator was + 13x.
The factors of 10 are switched in the binomials. The middle term would be 1x if this was correct.
This option is the solution to the numerator after it is factored.

Factor completely: $6 x^{3} - 6 x y^{2} + 9 x^{2} - 9 y^{2}$

$(6 x + 9) (x^{2} - y^{2})$
$3 (2 x + 3) (x + y) (x - y)$
$3 (2 x + 3) {(x + y)}^{2}$
cannot be factored

$3 (2 x^{3} - 2 x y^{2} + 3 x^{2} - 3 y^{2}) 3 (2 x (x^{2} - y^{2}) + 3 (x^{2} - y^{2})) 3 (2 x + 3) (x^{2} - y^{2}) 3 (2 x + 3) (x + y) (x - y)$

Note

The terms are not finished being factored.
The difference of two squares is not the same as a binomial squared.
This option is a factor by grouping with a GCF.

Use synthetic substitution to find $f (- 3)$ for $f (x) = x^{6} + 2 x^{5} - x^{4} + x^{3} - 10 x^{2} + 2$

–13 
1,073
47
–141

Note

This option is the answer if 0x is not included when dividing.
This option is f (3).
This option occurs if the remainder is multiplied by –3.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L1	L1	L3	L3	L5	L6	L7	L7	L1	L7	L3	L6

L = Lesson in this level, A1 = Algebra 1: Principles of Secondary Mathematics, FD = Foundational Knowledge