If there are real roots to be found for this polynomial, the Rational Root Theorem and synthetic division are the best way to find them. I teach from a book that uses c and d for the possible roots of the polynomial. C is our constant, 2, and d is the leading coefficient, 1. The factors of 2 are +/- 1 and +/-2. The factors for 1 are +/-1 only. Meaning, in all, there are 4 possibilities as roots for this polynomial. But there are only 3 total (because our polynomial is a third degree), so we have to find the first one, at least, from our possibilities above. Let's try x = -1, factor form (x + 1). If there is no remainder when we do the synthetic division, then -1 is a root. Put -1 outside the "box" and the coefficients from the polynomial inside: -1 (1 2 -1 -2). Bring down the first coefficient of 1 and multiply it by the -1 outside to get -1. Put that -1 up under the 2 and add to get 1. Multiply 1 times the -1 to get -1 and put that -1 up under the -1 and add to get -2. -1 times -2 is 2, and -2 + 2 = 0. So we have our first root of (x+1). The numbers we get when we do the addition along the way are the coefficients of our new polynomial, the depressed polynomial (NOT a sad one cuz it hates math, but a new polynomial that is one degree less than that of which we started!). The new polynomial is 
. That can also be factored to find the remaining 2 roots. Use standard factoring to find that the other 2 solutions are (x+2) and (x-1). Our solutions then are x = -2, -1, 1, choice B from above.
Enlarge the triangle by scale factor 0.5 using 3,1 as the centre of enlargement
1 answer:
You might be interested in
<h3>Answer: Choice A</h3>
The first line shown in choice A is which means "the first term is -2"
The next line in choice A means "the nth term () is found by multiplying the prior term (
) by 8". Put another way: multiply each term by 8 to get the next term.
first term = -2
second term = 8*(first term) = 8*(-2) = -16
third term = 8*(second term) = 8*(-16) = -128
fourth term = 8*(third term) = 8*(-128) = -1024
and so on.
