# Anyone how to do question 17 ii?

### 3 Answers

- ?Lv 75 months agoFavorite Answer
4x³ - 6x² + 1 = (x - 2)Q(x) + ax - 3

Subtract ax - 3 from both sides

4x³ - 6x² - ax + 1 = (x - 2)Q(x)

Divide both sides by (x - 2) but the final term must be -2 since -2 * -2 = 4

4x² + 2x - 2 remainder(-a + 6x) = Q(x)

Subtract the remainder from both sides

4x² + 2x - 2 = Q(x) + (a - 6)x

since the coefficient of x in the original term is zero

a - 6 = 0

a = 6 <–––––

Q(x) = 4x² + 2x - 2 <––––––

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Check

(x - 2)Q(x) + ax - 3 = 4x³ - 6x² + 1

(x - 2)(4x² + 2x - 2) + 6x - 3 = 4x³ - 6x² + 1

(4x³ + 2x² - 2x) + (- 8x² -4x + 4) + 6x - 3 = 4x³ - 6x² + 1

4x³ + (2x² - 8x²) + (-2x - 4x + 6x) + (4 - 3) = 4x³ - 6x² + 1

4x³ - 6x² + 0x + 1 = 4x³ - 6x² + 1 ✔︎

- 5 months ago
Thats the whole question. The value of A is 6 and thats all I have. So I dont really know how to move on to II frok this

- AlanLv 75 months ago
I would want to see the

rest of the question before answering that.

What you have by itself doesn't make any sense.

P(x) - 1 is divided by 3x^2 + 5

then what ?

with a remainder of what?

with no remainder?

Explain whether 3x^2 + 5 is a factor

P(x)-1 ?

No one can based on your information alone

[P(x)- 1]/ [3x^2 + 5] = (who knows ? with a remainder?

without a remainder? on a Tuesday?