The remainder theorem says that dividing a polynomial <em>f(x)</em> by a 1st-degree polynomial <em>g(x)</em> = <em>x</em> - <em>c</em> leaves a remainder of exactly <em>f(c)</em>.
(a) With <em>f(x)</em> = <em>px</em> ³ + 4<em>x</em> - 10 and <em>d(x)</em> = <em>x</em> + 3, we have a remainder of 5, so
<em>f</em> (-3) = <em>p</em> (-3)³ + 4(-3) - 10 = 5
Solve for <em>p</em> :
-27<em>p</em> - 12 - 10 = 5
-27<em>p</em> = 27
<em>p</em> = -1
(b) With <em>f(x)</em> = <em>x</em> + 3<em>x</em> ² - <em>px</em> + 4 and <em>d(x)</em> = <em>x</em> - 2, we have remainder 8, so
<em>f</em> (2) = 2 + 3(2)² - 2<em>p</em> + 4 = 8
-2<em>p</em> = -10
<em>p</em> = 5
(you should make sure that <em>f(x)</em> was written correctly, it's a bit odd that there are two <em>x</em> terms)
(c) <em>f(x)</em> = 2<em>x</em> ³ - 4<em>x</em> ² + 6<em>x</em> - <em>p</em>, <em>d(x)</em> = <em>x</em> - 2, <em>R</em> = <em>f</em> (2) = 18
<em>f</em> (2) = 2(2)³ - 4(2)² + 6(2) - <em>p</em> = 18
12 - <em>p</em> = 18
<em>p</em> = -6
The others are done in the same fashion. You would find
(d) <em>p</em> = 14
(e) <em>p</em> = -4359
(f) <em>p</em> = 10
(g) <em>p</em> = -13/2 … … assuming you meant <em>f(x)</em> = <em>x</em> ⁴ + <em>x</em> ³ + <em>px</em> ² + <em>x</em> + 20
7/28 as a decimal would be 0.28
Answer:
I don't understand the language
Step-by-step explanation:
am not form Paris
So the first one is correct which implies that the third one is also correct.
But these are not the only ones.
The last one is also correct.
Hope this helps.
r3t40
