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C2 - The Remainder Theorem HELP

Hi,

I'm doing these questions on the remainder theorem and need help:

1) a) Find the value of the constant a, such that, for all values of the constant b, one root of the equation 2x³ + ax + 4 = b(x-2) is equal to 2.

b) When a has this value, find the set of values of b for which the given equation has three distinct roots.

2) Given that f(x) = x² + px + q, determine the values of the constants p and q so that both

a) f(x) has a turning point when x=-3

b) the remainder when f(x) is divided by x+2 is 2
Show that, with these values of p and q, f(x) is greater than or equal to 1.

I would really appreciate it if someone could please help me, and explain how to work it out.

Thank you very much
Reply 1
Avatar for BCHL85
BCHL85
12
1) a) Substitue x = 2 to the function, you will get a.( a = -10)

b) You already know one root is 2, so factorise the function something like (x-2)(2x² ...) and find the b²-4ac > 0 to get condition for b. Also, you must find the condition that the others 2 roots must not be 2 as well.
Reply 2
Avatar for steve2005
steve2005
17
sweet_gurl
Hi,
1) a) Find the value of the constant a, such that, for all values of the constant b, one root of the equation 2x³ + ax + 4 = b(x-2) is equal to 2.

b) When a has this value, find the set of values of b for which the given equation has three distinct roots.

When b = - 4 there are only two roots. When b> -1 there are three roots.
Picture 10.png84.6KB
Reply 3
Avatar for steve2005
steve2005
17
sweet_gurl
Hi,
2) Given that f(x) = x² + px + q, determine the values of the constants p and q so that both

a) f(x) has a turning point when x=-3

b) the remainder when f(x) is divided by x+2 is 2
Show that, with these values of p and q, f(x) is greater than or equal to 1.
Thank you very much


Here is my solution.
Picture 12.png37.1KB
Reply 4
Avatar for steve2005
steve2005
17
steve2005
Here is my solution.


Here is a way of finding ' p ' without differentiation.
Picture 4.png75.0KB

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