What is the square root of 67 times 15? I get lost in the numbers.

Anna35 [415]

The best and most correct answer among the choices provided by your question is the second choice or letter B.

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7 0

2 years ago

Use synthetic division with the factor x + 1 to completely factor LaTeX: x^3+2x^2-5x-6x 3 + 2 x 2 âˆ’ 5 x âˆ’ 6.

notsponge [240]

Answer:

$x^3+2x^2-5x-6=\left(x+1\right)\left(x-2\right)\left(x+3\right)$ .

Step-by-step explanation:

To find $\frac{x^{3} + 2 x^{2} - 5 x - 6}{x + 1}$ using synthetic division you must:

Write the problem in a division-like format. To do this:

Take the constant term of the divisor with the opposite sign and write it to the left.

Write the coefficients of the dividend to the right.

$\begin{array}{c|cccc}&x^{3}&x^{2}&x^{1}&x^{0}\\-1&1&2&-5&-6\\&&\\\hline&\end{array}$

Step 1: Write down the first coefficient without changes:

$\begin{array}{c|rrrr}-1&1&2&-5&-6\\&&\\\hline&\1\end{array}$

Step 2:

Multiply the entry in the left part of the table by the last entry in the result row.

Add the obtained result to the next coefficient of the dividend, and write down the sum.

$\begin{array}{c|rrrr}-1&1&2&-5&-6\\&&\left(-1\right) \cdot 1=-1\\\hline&{1}&{2}+\left({-1}\right)={1}\end{array}$

Step 3:

Multiply the entry in the left part of the table by the last entry in the result row.

Add the obtained result to the next coefficient of the dividend, and write down the sum.

$\begin{array}{c|rrrr}{-1}&1&2&{-5}&-6\\&&-1&\left({-1}\right) \cdot {1}={-1}\\\hline&1&{1}&\left({-5}\right)+\left({-1}\right)={-6}\end{array}$

Step 4:

Multiply the entry in the left part of the table by the last entry in the result row.

Add the obtained result to the next coefficient of the dividend, and write down the sum.

$\begin{array}{c|rrrr}{-1}&1&2&-5&{-6}\\&&-1&-1&\left({-1}\right) \cdot \left({-6}\right)={6}\\\hline&1&1&{-6}&\left({-6}\right)+{6}={0}\end{array}$

We have completed the table and have obtained the following resulting coefficients: 1, 1, −6, 0.

All the coefficients except the last one are the coefficients of the quotient, the last coefficient is the remainder.

Thus, the quotient is $x^{2}+x-6$ , and the remainder is 0.

$\frac{x^{3} + 2 x^{2} - 5 x - 6}{x + 1}=x^{2} + x - 6+\frac{0}{x + 1}=x^{2} + x - 6$

Now, we factor $x^{2}+x-6$

$\left(x^2-2x\right)+\left(3x-6\right)\\x\left(x-2\right)+3\left(x-2\right)\\\left(x-2\right)\left(x+3\right)$

Therefore,

$x^3+2x^2-5x-6=\left(x+1\right)\left(x-2\right)\left(x+3\right)$

4 0

3 years ago

John's age increased by 11 is 5 less than 5 times his age. How old is he? If x represents John's age, which of the following equ

Arturiano [62]

Not entirely sure about the equation, but hes 4. 4 plus 11 is 15, fifteen plus 5 is 20. (fifteen is five less than 20) and 20 is 5 times his age, which is 4

7 0

3 years ago

Read 2 more answers

2(4x+9)=8x+18 solve for x

Olegator [25]

2(4x+9)=8x+18

8x+18=8x+18

8x=8x

x=x

8 0

2 years ago

Read 2 more answers