Ask question

Ask question

All categories

3 years ago

11

Consider the polynomial 16x2 – 1. What is the value of ac? What is the value of b? What two numbers multiply to get ac and add t

o get b? The factored form of 16x2 – 1 is

2 answers:

WINSTONCH [101]3 years ago

9 0

ANSWER

The given expression is
$16 {x}^{2} - 1$

This expression can be rewritten as,

$16 {x}^{2} + 0x - 1$

By comparing to
$a{x}^{2} + bx + c$
$a=16,b=0,c=-1$
$ac = - 16$

Two numbers that we multiply to get
$- 16$
and add to get zero are
$- 4 \: and \: 4$

The factored form is

$16 {x}^{2} - 4x + 4x- 1$

$4x(4x - 1) +1( 4x- 1)$

$(4x + 1)(4x - 1)$

We factor further to get,

Sphinxa [80]3 years ago

5 0

1. -16

2. 0

3. -4 and 4

4. (4x-1)(4x+1)

You might be interested in

Emily is saving $300 a month for a $9,000 down payment on a new car. she will have her down payment total in 18 months. she got

Burka [1]

It would fewer her months by 10

4 0

4 years ago

Read 2 more answers

Can you please answer this question?

liraira [26]

300miles/20sec=?miles/1min
900miles/60sec=?miles/1min
900miles/1min=?miles/min
?=900 miles

3 0

3 years ago

Read 2 more answers

Which strategy best explains how to solve this problem? Rebecca bought a meal and two snacks for lunch each day. The meal cost $

Fittoniya [83]

Given:
Cost of lunch per day = 1 meal and 2 snacks

C = 5.5 + 2(0.75) = 5.5 + 1.5 = 7

7 * 12 days = 84

Based on the choices, the best strategy would be:

<span> A. Make a table. Write the numbers 1 to 12 in the top row of the table (the number of days). In the first box on the second row, write $7. This is how much Rebecca spends in 1 day. In each of the next boxes in the second row, write the amount Rebecca spends by adding $7 to the previous amount. The answer in box 12 is the total amount Rebecca spent after 12 days.</span>

6 0

3 years ago

PLLLLLEASE I NEED HELP SO BADLY PLEASE ASAP PLEASE PLEASE T_T

olchik [2.2K]

Answer:

$x^2+7x-8-\frac{3}{x-7}$

Step-by-step explanation:

I think this question is a bit hard to explain by typing, so I'll add an image down below.

I used long division for this question, and the final answer is: $x^2+7x-8-\frac{3}{x-7}$

4 0

2 years ago

Which summation formula represents the series below? 1 + 2 + 6 + 24

krek1111 [17]

Question:

Which summation formula represents the series below? 1 + 2 + 6 + 24

(a) $\sum_{n=2}^{5}(n-1) !$

(b) $\sum_{n=0}^{3} n !$

(c) $\sum_{n=1}^{4}(n+1) !$

(d) $\sum_{n=2}^{5} n !$

Answer:

Option a: $\sum_{n=2}^{5}(n-1) !$ is the correct answer.

Explanation:

Option a: $\sum_{n=2}^{5}(n-1) !$

By substituting the values of n and expanding the summation, we have,

$(2-1) !+(3-1) !+(4-1) !+(5-1) !$

Subtracting, we have,

$1 !+2!+3 !+4 !$

Expanding the factorial,

$1+(2*1)+(3*2*1)+(4*3*2*1)$

Simplifying, we get,

$1+2+6+24$

Thus, the summation $\sum_{n=2}^{5}(n-1) !$ represents the series $1+2+6+24$

Hence, Option a is the correct answer.

Option b: $\sum_{n=0}^{3} n !$

By substituting the values of n and expanding the summation, we have,

$0!+1!+2!+3!$

Expanding the factorial,

$0+1+(2*1)+(3*2*1)$

Simplifying, we get,

$0+1+2+6$

Thus, the summation $\sum_{n=0}^{3} n !$ does not represents the series $1+2+6+24$

Hence, Option b is not the correct answer.

Option c: $\sum_{n=1}^{4}(n+1) !$

By substituting the values of n and expanding the summation, we have,

$(1+1) !+(2+1) !+(3+1) !+(4+1) !$

Adding, we have,

$2!+3!+4!+5!$

Expanding the factorial,

$(2*1)+(3*2*1)+(4*3*2*1)+(5*4*3*2*1)$

Simplifying, we get,

$2+6+24+120$

Thus, the summation $\sum_{n=1}^{4}(n+1) !$ does not represents the series $1+2+6+24$

Hence, Option c is not the correct answer.

Option d: $\sum_{n=2}^{5} n !$

By substituting the values of n and expanding the summation, we have,

$2!+3!+4!+5!$

Expanding the factorial,

$(2*1)+(3*2*1)+(4*3*2*1)+(5*4*3*2*1)$

Simplifying, we get,

$2+6+24+120$

Thus, the summation $\sum_{n=2}^{5} n !$ does not represents the series $1+2+6+24$

Hence, Option d is not the correct answer.

Hence, the correct answer is Option a: $\sum_{n=2}^{5}(n-1) !$

6 0

3 years ago