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3 years ago

What is the quotient (2x4 – 3x3 – 3x2 + 7x – 3) ÷ (x2 – 2x + 1)?

Mathematics

2 answers:

MrRa [10]3 years ago

8 0

Answer:

Quotient: $2x^2+x-3$

Please see the attachment.

Step-by-step explanation:

Given: $(2x^4-3x^3-3x^2+7x-3)\div (x^2-2x+1)$

We are given rational expression and need to find quotient.

Using long division method to find the quotient.

First we get rid of $2x^4$ by $x^2$

$x^2-2x+1$ ) $2x^4-3x^3-3x^2+7x-3$ ( $2x^2+x-3$

$-2x^4+4x^3-2x^2$

$x^3-5x^2+7x$

$-x^3+2x^2-x$

$-3x^2+6x-3$

$3x^2-6x+3$

$0$

Hence, The quotient of division is $2x^2+x-3$

Nat2105 [25]3 years ago

5 0

Here you have two polynomials:

1. The dividend - $f(x)=2x^4 - 3x^3 - 3x^2 + 7x - 3$

2. The divisor - $g(x)=x^2 - 2x + 1$ .

Since divisor is perfect square $g(x)=x^2 - 2x + 1=(x-1)^2$ , you should check what is the quotient after division f(x) by (x-1):

$f(x)=2x^4 - 3x^3 - 3x^2 + 7x - 3=(x-1)(2x^3-x^2-4x+3)=(x-1)(x-1)(2x^2+x-3)=(x-1)^2(2x^2+x-3)=g(x)(2x^2+x-3).$

Then the quotient is $2x^2+x-3$ .

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Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is