The expression which represents the other factor, or factors, of the given polynomial is option (C) (2x-1)(x+1)
A cubic equation in algebra is a one-variable equation of the form ax3+bx2+cx+d=0 where an is nonzero. The roots of the cubic function defined by the left side of this equation are the solutions to this equation.
Given expression 2x³-3x²-3x+2 whose one of factor is (x-2)
We have to find second factor of given equation
First we will be rational root theorem to given expression so will get following expression:

So one factor is (x-1) and now simplifying
we get 2x² - 5x +2 and the factor of 2x² - 5x +2 will be (2x-1)(x-2)
Hence the expression which represents the other factor, or factors, of the given polynomial is option (C) (2x-1)(x+1)
Learn more about Polynomial here:
brainly.com/question/4142886
#SPJ10
Answer:
okay it simple the answer is D
Step-by-step explanation:
im just smart I guess
1. The figure is a prism: V = lwh = (12.1)(5.3)(4.2) = 269.3 cu. cm.
2. The figure is also a prism but the base is triangular: V = 1/2*b*h*l = 1/2*(13)*(sqrt(3)/2*13)*24 = 1756.3 cu. yd.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
Answer:
Answer is Option 2
Step-by-step explanation:
Triangles equal to 180 degrees in total. So it is 55+54+x+74=180
The simplified version that I just wrote is Option 2
Answer:
<u>Given function</u>
#15 Find the inverse of h(x)
<u>Substitute x with y and h(x) with x and solve for y:</u>
- x = 2y - 1
- 2y = x + 1
- y = 1/2x + 1/2
<u>The inverse is:</u>
#16 The graph with both lines is attached.
The x- and y-intercepts of both functions have reversed values.
#17 Table of the inverse function will contain same numbers with swapped domain and range.
<u>Initial look is like this:</u>
- <u>x | -3 | -2 | -1 | 0 | 1 | 2 | 3</u>
- h⁻¹(x) | -1 | | 0 | | 1 | | 2
<u>The rest of the table is filled in by finding the values:</u>
- <u>x | -3 | -2 | -1 | 0 | 1 | 2 | 3</u>
- h⁻¹(x) | -1 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2