So, we would need to remember, the one way that me personally would view square rooting would be by simplifying them, and that number would go into that number that many times. So, when doing this kind of problem, we are not truly going to do this, but we are just going to simplify it, and to see what other square "rooter" would go into that.
So, we would need to remember a (key) point, <em>we aren't just multiplying, for the most part, we're simplifying. </em>
Our result:
![\boxed{\boxed{\bf{2a^2b \sqrt[4]{24a^2b^3} }}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5Cbf%7B2a%5E2b%20%5Csqrt%5B4%5D%7B24a%5E2b%5E3%7D%20%7D%7D%7D)
We didn't just multiplied it, we also simplified it also.
Answer:
95cm
Step-by-step explanation:
45cm is the initial height
each year grow 10cm
growing time 5 years
45+(10 × 5)=95 cm
B. The total area painted is 864; they must buy two cans of paint.
Step-by-step explanation:
Step 1:
A rectangle's area can be calculated by multiplying its length and its width. The wall is made up of 5 different types of rectangular walls. All walls are 8 feet tall but the length varies.
Step 2:
The area of the 20 feet long wall = 
The area of the 10 feet long wall = 
The area of the 5 feet long wall = 
The area of the 4 feet long wall = 
The area of the 15 feet long wall = 
The area of all the walls = 432 square feet.
Since there are two sides for every wall, total area = 
square feet.
Step 3:
If one paint can covers 500 square feet,
the number of cans required to paint 864 square feet = 
cans.
so 2 paint cans are needed to paint 864 square feet which is option B.
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.
