Core 2 Maths - Factor/Remainder Theorem

Struggling on question 8 shown in the images. I understandly the concept and the way I have worked through it is correct, but my answer seems to be slightly different from the mark scheme. Any help would be appreciated.

In the fourth line of your writing, it should be just a and not a^2.

In the expression ax^2 only the x is squared and not a.

You made the same mistake for x = -2

(edited 9 years ago)

Reply 2

Parallex

6

let (x^3 + ax^2+bx-6 = 0) = f(x)[br]if (x-1) and (x+2) are factors then (x+3) must also be a factor as -1*2*3 = -6 [br](x-1)(x+2) = x^2+x-2[br]so f(x) = (x^2+x-2)(x+3)[br][br]comparing coefficients we get:[br][br]x^2 terms: [br](x^2)(3) = 3x^2[br](x)(x) = x^2[br]3x^2 + x^2 = 4x^2[br][br]x terms:[br](x)(-2) = -2x[br](x)(3) = 3x[br]-2x + 3x = x[br][br]therefore, a = 4 and b = 1[br][br]

I am so bad with LaTeX.

(edited 9 years ago)

Reply 3

Parallex

6

Forgot that I wasn't supposed to post the answer but you had the answer already so I suppose that's alright. Your method was pretty long winded, realising (x+3) is the final factor and working from there would've been quicker. ^^

Reply 4

OL350

OP

2

Thanks for the help guys, really appreictaed.
In response to your comment made, Parallex, I really do understand my method might be a little longer than yours, however it's one that is easily understandable. I seem to lose yours after the 'comparing coefficients we get' statement - any help explaining it would great.
Thanks for the short and helpful tip 123ab. However, when applying this knowledge to another (almost identical) question, for some reason it doesn't seem to work...I get two thirds for B, but the actual answer for both A and B is zero.

At the end of the fourth line of your writing, you wrote - 1 + 6 = 0 when it should be - (-1) + 6 = 0

Remember, x is -1 and so - x is

- (-1) which is + 1

(edited 9 years ago)

Reply 6

Parallex

6

Original post by OL350

Thanks for the help guys, really appreictaed.
In response to your comment made, Parallex, I really do understand my method might be a little longer than yours, however it's one that is easily understandable. I seem to lose yours after the 'comparing coefficients we get' statement - any help explaining it would great.
Thanks for the short and helpful tip 123ab. However, when applying this knowledge to another (almost identical) question, for some reason it doesn't seem to work...I get two thirds for B, but the actual answer for both A and B is zero.

So when we compare coefficients, we're looking at which terms pair up on the LHS (left hand side) to match terms on the RHS (right hand side).

so we have

f(x) =x^3 + ax^2 + bx - 6 = 0

we have also found out the three factors of the polynomial:

(x-1) (x+2) (x+3)

we could just multiply these out.

(x-1)(x+2)= x^2+x-2[br][br](x^2+x-2)(x+3) = x^3+x^2-2x+3x^2+3x-6

tidying this up we get

x^3+4x^2+x-6

so we have found that

f(x) = x^3+4x^2+x-6

a is the

x^2

coefficient whilst b is the

x

coefficient. Sorry if it was a bit hard to understand. :smile:

Reply 7

Don Pedro K.

9

Hey there,

I worked this question out using simultaneous equations; the whole thing took me like 3-4 minutes so it's not time consuming at all and it's logical :smile:

I would post a picture of my working but my camera is really bad so I can't haha! I see you tried to use simultaneous equations also, but you went wrong early so have another look (hint: the a doesn't get squared anywhere) Good luck!

Reply 8

OL350

OP

2

Any help on question 2D would be appreciated.

OP

Original post by OL350

Any help on question 2D would be appreciated.

I think youve wrote the question out wrong 2d doesnt say divide by (2x+3)

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Reply 11

OL350

OP

2

Haha haha

...my bad. I always seem to make that mistake! Thanks for the heads up, been trying to work it out for ages!

Reply 12

Covington4

19

Haha thats alright I do it all the time!

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Reply 13

OL350

OP

2

The answer book states C is equal to -2. Surely that is wrong...(-1) X (1) X (2) equals positive 2.

Reply 14

Covington4

19

Original post by OL350

The answer book states C is equal to -2. Surely that is wrong...(-1) X (1) X (2) equals positive 2.

-1 x 1 = -1
-1 x 2 = -2

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Reply 15

OL350

OP

2

Sorry I have written my post above wrong... (-1) X (1) X (-2) equals positive 2, however the book gives -2 as the answer?

Reply 16

lukejoshjames

7

What value did you get for A and B?

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Reply 17

OL350

OP

2

Please could someone explain the binomial expansion question in the photo. I understand how to do each individual expansion but I'm not sure on how to get a final answer.

Reply 18

Covington4

19

Original post by OL350

Please could someone explain the binomial expansion question in the photo. I understand how to do each individual expansion but I'm not sure on how to get a final answer.