Author Topic: Functions (Read 8064 times) Tweet Share

lanvins · « **Reply #30 on:** February 25, 2008, 08:59:00 pm »

More questions

1.how do you divide x^4-9x^3+25x^2-8x-2 by x^2-2?

2. Find the value of a for which (1-2a)x^2+5ax+(a-1)(a-8) is divisible by (x-2) but not by (x-1)

3.i. Prove that the expression x^3+(k-1)x^2+(K-9)x-7 is divisible by x+1 for all values of k

ii. Find the value of k for which the expression has a remainder of 12 when divided by x-2

4. The expression 4x^3 +ax^2 -5x +b leaves remainder of -8 and 10 when divided by (2x-3) and (x-3) repectively. Calculate the values of a and b

Toothpaste · « **Reply #31 on:** February 25, 2008, 10:05:17 pm »

Quote from: lanvins on February 25, 2008, 08:59:00 pm

4. The expression 4x^3 +ax^2 -5x +b leaves remainder of -8 and 10 when divided by (2x-3) and (x-3) repectively. Calculate the values of a and b

Let $P(x) = 4x^3 +ax^2 -5x +b$

(2x-3) leaves a remainder of -8, so...

$P(\frac{3}{2}) = 4(\frac{3}{2} )^3 +a(\frac{3}{2} )^2 -5(\frac{3}{2}) +b = -8$

$\implies \frac{27}{2} + \frac{9}{4}a - \frac{15}{2} +b = -8$

$\implies 6 + \frac{9}{4}a + b = -8$

$\implies \frac{9}{4}a +b = -14$ ... (i)

(x-3) leaves a remainder of 10, so...

$P(3) = 4(3)^3 +a(3)^2 -5(3) +b = 10$

$\implies 108 + 9a -15 +b = 10$

$\implies 9a +b = -83$

$\implies b = -83 - 9a$ ... (ii)

substitute (ii) into (i):

$\frac{9}{4}a + (-83 - 9a) = -14$

$\frac{-27}{4}a = 69$

$\therefore a = \frac{-92}{9}$

$\therefore b = 9$

Does this correlate to the answers?

lanvins · « **Reply #32 on:** February 25, 2008, 10:13:11 pm »

yep that's right, thanks

Collin Li · « **Reply #33 on:** February 25, 2008, 10:37:36 pm »

Question 1 should just be done by simple long division (or some type of short division).

Question 2
Find the value of a for which $(1-2a)x^2+5ax+(a-1)(a-8)$ is divisible by $(x-2)$ but not by $(x-1)$

Let $P(x) = (1-2a)x^2+5ax+(a-1)(a-8)$

If $(x-2)$ is a factor (i.e.: divisible), then $P(2) = 0$

$4(1-2a)+10a+(a-1)(a-8) = 0$

$\implies 4 - 8a + 10a + (a^2 - 9a + 8) = 0$

$\implies a^2 - 7a + 12 = 0$

$\implies (a-4)(a-3) = 0$

$\implies a = 3,\, a = 4$

But so that $(x-1)$ is not a factor, $P(1) \neq 0$

$(1-2a)+5a+(a-1)(a-8) \neq 0$

$\implies 1 - 2a + 5a + (a^2 - 9a + 8) \neq 0$

$\implies a^2 - 6a + 9 \neq 0$

$\implies (a-3)^2 \neq 0$

$\implies a \neq 3$

Therefore, $a = 4$

Toothpaste · « **Reply #34 on:** February 25, 2008, 10:45:26 pm »

Starting Q1 off for you...

$\begin{equation*} \begin{array}{rc@{}c} & \multicolumn{2}{l}{\, \, \, x^2 -9x +23} \vspace*{0.12cm} \\ \cline{2-3} \multicolumn{1}{r}{x^2 - 2 \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} x^4-9x^3+25x^2-8x-2} \\ & \multicolumn{2}{l}{\, \, \, x^4+0 -2x^3} \vspace*{0.12cm} \\ \cline{2-3} & \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}} -9x^3+23x^2-8x-2} \\ & \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}} -9x^3+0x^2+18x} \vspace*{0.12cm} \\ \cline{2-3} & \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}...{}-{}} etc\dotsb} \end{array} \end{equation*} $

remainder of $10x - 48$

$\therefore \frac{x^4-9x^3+25x^2-8x-2}{x^2-2}$

$= x^2-9x+23$ $+ \frac{10x-48}{x^2-x}$

Collin Li · « **Reply #35 on:** February 25, 2008, 10:45:46 pm »

Question 3

i. Prove that the expression $x^3+(k-1)x^2+(k-9)x-7$ is divisible by $x+1$ for all values of $k$

Let $P(x) = x^3+(k-1)x^2+(k-9)x-7$

Divisible by $x+1 \implies P(-1) = 0$ :

Starting from the LHS:

$P(-1) = (-1)^3 + (k-1)(-1)^2 + (k-9)(-1) - 7$

$= -1 + k - 1 - k + 9 - 7$

$= -9 + 9 + k - k$

$= 0$ (as required)

ii. Find the value of $k$ for which the expression has a remainder of 12 when divided by $x-2$

A remainder of 12 when divided by $x-2 \implies P(2) = 12$ :

$12 = 8 + 4(k-1) + 2(k-9) - 7$

$\implies 12 = 8 + 4k - 4 + 2k - 18 - 7$

$\implies 33 = 6k$

$\implies k = 5.5$

lanvins · « **Reply #36 on:** February 26, 2008, 05:47:30 pm »

yea that makes sense, thanks

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