Remainder Theorem question

a. Find the value of a for which (x+2) is a factor of

2x^3+ax^2+5x+6

b. Show that, for this value of a, the cubic equation

2x^3+ax^2+5x+6=0

has only one root.

I understand part a well and got the correct answer of a=5.

However, I'm stuck on part b. Below's my working out:

2x^3+5x^2+5x+6

(x+2)(2x^2+x+3)

I now don't understand what to do as there is (x+2) included. So what do I do to fully answer part b???? The whole equation is cubic, not quadratic.

Thank you very much.

(edited 11 years ago)

Reply 1

msmith2512

1

If (x+2)(2x^2+x+3) =0 what can be said about the two factors? If you know this how many values of x satisfy the equation?

Reply 2

L'Evil Fish

18

See how your other brackets only.has a certain type of root?

Use.the discriminant and see if the answer is below 0 :smile:

Reply 3

krisshP

OP

14

Original post by msmith2512

If (x+2)(2x^2+x+3) =0 what can be said about the two factors? If you know this how many values of x satisfy the equation?

One's a quadratic and the other I think is perhaps linear.

Wait, we can draw up that one root is clearly x=-2. The quadratic equation 2x^2+x+3 has a negative discriminant of -23. Hence there are no sultions to 2x^2+x+3. However, as I just said, x=-2 is a solution. In fact this is the only solution.

Am I right or not????
Thanks

Reply 4

L'Evil Fish

18

Original post by krisshP

One's a quadratic and the other I think is perhaps linear.

Wait, we can draw up that one root is clearly x=-2. The quadratic equation 2x^2+x+3 has a negative discriminant of -23. Hence there are no sultions to 2x^2+x+3. However, as I just said, x=-2 is a solution. In fact this is the only solution.

Am I right or not????
Thanks

As the other roots are imaginary :smile:

Reply 5

krisshP

OP

14

Original post by shadab786ahmed

As the other roots are imaginary :smile:

Imaginery??????????? What do you mean??? Is that F-maths stuff?

btw, thanks for helping me out. :smile:

Reply 6

krisshP

OP

14

it's just that I don't do Further maths

Reply 7

Notnek

21

Original post by krisshP

Imaginery??????????? What do you mean??? Is that F-maths stuff?

btw, thanks for helping me out. :smile:

Yes it's further maths stuff. If you call

\sqrt{-1}=i

then you can use the quadratic formula to express solutions that "don't exist" in terms of

i

which is the imaginary unit.

At A Level, you just need to say that solutions that arise from a negative discriminant do not exist. So for this question, the equation has one root.

(edited 11 years ago)

Reply 8

krisshP

OP

14

Original post by notnek

Yes it's further maths stuff. If you call

\sqrt{-i}=i

then you can use the quadratic formula to express solutions that "don't exist" in terms of

i

which is the imaginary unit.

At A Level, you just need to say that solutions that arise from a negative discriminant do not exist. So for this question, the equation has one root.