It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if
</span><span>
</span><span>In notation we write respectively
</span>
Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²<span>
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c</span>²
That is to say, if c = -2, f(x) is continuous at x = 4.
Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers 
<span><span><span><span>2<span>x^3</span></span>+<span>6x</span></span>+152
</span><span>x+4</span></span><span>=<span><span><span><span>2<span>x^3</span></span>+<span>6x</span></span>+152
</span><span>x+4</span></span></span><span>=<span><span><span>2<span>(<span>x+4</span>)</span></span><span>(<span><span><span>x^2</span>−<span>4x</span></span>+19</span>)
</span></span><span>x+4</span></span></span><span>=<span><span><span>2<span>x^2</span></span>−<span>8x</span></span>+<span>38= is the answer</span></span></span>
well, the remainder theorem says that if the polynomial f(x) has a factor of (x-a), then if we just plug in the "a" in f(x) it'll gives a remainder, assuming (x-a) is indeed a factor, then that remainder must be 0, so if f(a) = 0 then indeed (x-a) is a factor of f(x). After all that mumble jumble, let's proceed, we have (x+1), that means [ x - (-1) ], so if we plug in -1 in f(x), we should get 0, or f(-1) = 0, let's see if that's true.
Answer:
60 %
Step-by-step explanation:
multiply .90 * 150
