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UkoKoshka [18]
3 years ago
13

What is the factored form of x^3-1

Mathematics
1 answer:
polet [3.4K]3 years ago
7 0

Answer:

the factors of x^3-1 are x^3 -1 = (x-1) (x^2 +x+1)\\

Step-by-step explanation:

We need to find factors of x^3 -1

x^3-1 \\(x)^3 -(1)^3\\We \,\, know\,\, a^3 -b^3 = (a-b) (a^2 +ab + y^2) \\So, \,\, using\,\, the \,\, formula\,\,\\Here\,\, a= x \,\,and\,\, b= 1\\(x)^3 -(1)^3 = (x-1) (x^2 +x+1)\\

So, the factors of x^3-1 are x^3 -1 = (x-1) (x^2 +x+1)\\

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Lucia draws a square and plots the center of the square. (image)
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3 years ago
What number must you add to complete the square?<br> x^2 + 2x = -1
Andru [333]
\large\begin{array}{l} \textsf{You must add 1 to complete the square.}\\\\ \textsf{There is a very common special product: (square of a sum)}\\\\ \mathsf{(a+b)^2=a^2+2ab+b^2}\\\\\\ \textsf{Take the given equation:}\\\\ \mathsf{x^2+2x=-1\qquad(i)}\\\\\\ \textsf{The first term is a square (x squared)}\\\\ \textsf{The second term is a product of 2, x and 1.} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{We need to find a proper value for b, so that if we add it to}\\\textsf{the left side of the equation, we get a perfect square.}\\\\ \textsf{So,}\\\\ \begin{array}{cccccccccc} \mathsf{a^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{a}&\!\!\!\cdot\!\!\!&\mathsf{b}&\!\!\!+\!\!\!&\mathsf{b^2}&=\mathsf{(a+b)^2}\\\\ \mathsf{\downarrow}&&&\!\!\!\!\!&\downarrow&\!\!\!\!\!&\downarrow&&\downarrow\\\\ \mathsf{x^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{x}&\!\!\!\cdot\!\!\!&\mathsf{1}&\!\!\!+\!\!\!&\mathsf{1^2}&=\mathsf{(x+1)^2} \end{array}\\\\\\ \textsf{Just by comparing the expressions above, we can conclude that}\\\textsf{b must be equal to 1, so we get a perfect square:}\\\\ \mathsf{(x+1)^2=x^2+2x+1} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{Back to (i), adding 1 to both sides:}\\\\ \mathsf{x^2+2x+1=-1+1}\\\\ \boxed{\begin{array}{c} \mathsf{(x+1)^2=0} \end{array}} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{If you want to solve this equation for x, you will find}\\\\ \mathsf{x+1=0}\\\\ \mathsf{x=-1}\quad\longleftarrow\quad\textsf{(solution)} \end{array}


If you're having problems understanding the answer, try seeing it through your browser: brainly.com/question/2112248


\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best regards! :-)} \end{array}

6 0
3 years ago
Read 2 more answers
7a=4(2a+11)-a <br> Please help me solve this!
Pepsi [2]
7a=4(2a+11)-a
7a=7a+44
7a-7a=7a+44-7a
0=44
no solutions
4 0
3 years ago
Help solving this math problem
daser333 [38]

Diagram of two triangles

The two triangles are similar because they are both right triangles, meaning that one angle is 90° and the other two are acute (less than 90°).

The diagram on the left is missing its hypotenuse - the variable <em>c</em><em>.</em><em> </em>The hypotenuse is opposite of the right angle. The diagram on the right side is missing one of its legs.

Note: The hypotenuse is the <em>longest</em><em> </em>side of a right triangle.

Word Problen

The legs are 21 blocks and 20 blocks because they are by the right angle. Use the Pythagorean theorem to find the diagonal path's length, which is the hypotenuse.

{a}^{2}  +  {b}^{2}  =  {c}^{2}

Standard form of Pythagorean theorem.

{21}^{2}  +  {20}^{2}  =  {c}^{2}

Equation with the legs substituted and the missing hypotenuse value - <em> </em><em>c</em><em>.</em>

841 =  {c}^{2}

Square the legs and add.

\sqrt{841}  = c

29 = c

Take the square root and simplify. The square root of 841 is 29 and -29, but distance is positive.

Thus the diagonal distance is 29 blocks.

Check by substituting.

5 0
3 years ago
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