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Simplify the expression (x^3-5x^2+7x-12)/(x-4) using long division.

Show your work

Polynomial long division:

\(\begin{array}{|rcll|} \hline \dfrac{x^3-5x^2+7x-12}{x-4} &=& x^2-x+3 \\ \hline \end{array}\)

exponential problem

4^(x) -2^(x+2) +4=0

\(\begin{array}{|rcll|} \hline 4^{x} -2^{x+2} +4 &=& 0 \\ (2\cdot2)^{x} -2^x\cdot2^2 +2^2 &=& 0 \\ (2^2)^{x} -2^x\cdot2^2 +2^2 &=& 0 \quad & | \quad (2^2)^{x} = 2^{2x} = 2^{x\cdot 2}= (2^x)^2 \\ (2^x)^2 -2^x\cdot2\cdot 2 +2^2 &=& 0 \quad & | \quad \text{binom } \\ (2^x-2)^2 &=& 0 \quad & | \quad \text{square root both sides}\\ 2^x-2 &=& 0 \quad & | \quad +2 \\ 2^x &=& 2 \quad & | \quad 2=2^1 \\ 2^x &=& 2^1 \\ \mathbf{x} &\mathbf{=}& \mathbf{ 1} \\ \hline \end{array}\)

Find the value(s) of m in the quadratic x²+x+m+1=0
such that there is only one solution.

\(\begin{array}{|lrcll|} \hline & x^2+1\cdot x+ (m+1) &=& 0 \\ & \text{ or } (x-x_1) (x-x_2) &=& 0 \quad & | \quad \text{one solution so } x_1 = x_2 \\ & (x-x_1) (x-x_1) &=& 0 \\ & (x-x_1)^2 &=& 0 \\ & x^2 \underbrace{-2x_1}_{=1}x +\underbrace{ x_1^2}_{=m+1} &=& 0 \quad & | \quad \text{compare the coefficients} \\\\ (1) & -2x_1 &=& 1 \\ & x_1 &=& -\frac{1}{2} \\\\ (2) & x_1^2 &=& m+1 \\ & \left(-\frac{1}{2}\right)^2 &=& m+1 \\ & \frac{1}{4} &=& m+1 \\ & \frac{1}{4}-1 &=& m \\ & \frac{1}{4}-\frac{4}{4} &=& m \\\\ &\mathbf{ -\frac{3}{4}} &\mathbf{=}& \mathbf{m} \\ \hline \end{array}\)

Expand and simplify.

(t-1)^3 + 6(t-1)^2 + 12(t-1) + 8

Formula:
\([1+(t-1)]^3 = t^3 = (t-1)^3 + 3(t-1)^2+3(t-1) + 1 \)

\(\begin{array}{|rcll|} \hline && \mathbf{(t-1)^3 + 6(t-1)^2 + 12(t-1) + 8 } \\ &=& (t-1)^3 + 3(t-1)^2+3(t-1) +1 \\ && \quad + 3(t-1)^2+9(t-1)+7 \\ &=& t^3 + 3(t-1)^2+9(t-1)+7 \\ &=& t^3 + 3(t^2-2t+1)+9t-9+7 \\ &=& t^3 + 3t^2-6t+3+9t-9+7 \\ &=& t^3 + 3t^2+3t+1 \quad & | \quad (1+t)^3 = t^3+3t^2+3t+1 \\ &\mathbf{=}& \mathbf{(1+t)^3} \\ \hline \end{array} \)

There is a row of Pascal's triangle that has three successive positive entries,"a" "b"  and "c"
such that "b" is double "c"
and "a" is triple "c"
If this row begins "1,n,"  then find n.

which term is the number 78987 in this sequence?

Here 0 is the 1st term:

The position of a palindrome within the sequence can be determined almost without calculation:

If the palindrome has an even number of digits,

prepend a 1 to the front half of the palindrome's digits.

Examples: 98766789=a(19876)

If the number of digits is odd, prepend the value of front digit + 1 to the digits from position 2 ... central digit.

Examples: 515=a(61), 8206028=a(9206), 9230329=a(10230).

Express a4(b - c) + b4 (c - a) + c4 (a - b) as the product of four factors.

(b-c),(c-a) and (a-b) are products of this expression

\(\begin{array}{|rcll|} \hline a^4(b-c)+b^4(c-a)+c^4(a-b) &=& (b-c)(c-a)(a-b)\cdot x \\\\ x &=& \dfrac{a^4(b-c)+b^4(c-a)+c^4(a-b)}{(b-c)(c-a)(a-b)} \\\\ &=& \dfrac{a^4b-a^4c+b^4c-b^4a+c^4a-c^4b}{-a^2b+a^2c-b^2c+b^2a-c^2a+c^2b} \\\\ &=& \dfrac{ a^4·b - a^4·c - a·b^4 + a·c^4 + b^4·c - b·c^4 }{-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2} \\\\ \hline \end{array}\)

Polynom division after variable c:

\(\small{ \begin{array}{|rcll|} \hline && \text {in divisor term with max power of c is: } -ac^2 \\ && \text {current residue: } a^4·b - a^4·c - a·b^4 + a·c^4 + b^4·c - b·c^4 \\ && \text {in current residue term with max power of c: } ac^4 \\ && \text {quotient } \frac{ac^4}{-ac^2} = -c^2 \\ && \text {product } -c^2·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^2·b·c^2 - a^2·c^3 - a·b^2·c^2 + a·c^4 + b^2·c^3 - b·c^4 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } a^4·b - a^4·c - a^2·b·c^2 + a^2·c^3 - a·b^4 + a·b^2·c^2 + b^4·c - b^2·c^3 \\ && \text {in current residue term with next lower power of c is: } a^2·c^3 \\ && \text {quotient } \frac{a^2·c^3}{-ac^2} = -a·c \\ && \text {product } -a·c·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^3·b·c - a^3·c^2 - a^2·b^2·c + a^2·c^3 + a·b^2·c^2 - a·b·c^3 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } a^4·b - a^4·c - a^3·b·c + a^3·c^2 + a^2·b^2·c - a^2·b·c^2 - a·b^4 + a·b·c^3 + b^4·c - b^2·c^3 \\ && \text {in current residue term with next lower power of c is: } a·b·c^3 \\ && \text {quotient } \frac{a·b·c^3}{-a·c^2} = -b·c \\ && \text {product } -b·c·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^2·b^2·c - a^2·b·c^2 - a·b^3·c + a·b·c^3 + b^3·c^2 - b^2·c^3 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } a^4·b - a^4·c - a^3·b·c + a^3·c^2 - a·b^4 + a·b^3·c + b^4·c - b^3·c^2 \\ && \text {in current residue term with next lower power of c is: } a^3·c^2 \\ && \text {quotient } \frac{a^3·c^2}{-a·c^2} = -a^2 \\ && \text {product } -a^2·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^4·b - a^4·c - a^3·b^2 + a^3·c^2 + a^2·b^2·c - a^2·b·c^2 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } a^3·b^2 - a^3·b·c - a^2·b^2·c + a^2·b·c^2 - a·b^4 + a·b^3·c + b^4·c - b^3·c^2 \\ && \text {in current residue term with next lower power of c is: } a^2·b·c^2 \\ && \text {quotient } \frac{a^2·b·c^2}{-a·c^2} = -a·b \\ && \text {product } -a·b·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^3·b^2 - a^3·b·c - a^2·b^3 + a^2·b·c^2 + a·b^3·c - a·b^2·c^2 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } a^2·b^3 - a^2·b^2·c - a·b^4 + a·b^2·c^2 + b^4·c - b^3·c^2 \\ && \text {in current residue term with next lower power of c is: } a·b^2·c^2 \\ && \text {quotient } \frac{a·b^2·c^2}{-a·c^2} = -b^2 \\ && \text {product } -b^2·(-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2) = a^2·b^3 - a^2·b^2·c - a·b^4 + a·b^2·c^2 + b^4·c - b^3·c^2 \\ && \text {subtract product form current residue: } \\ && \text {current residue: } 0 \\\\ x&=& \dfrac{a^4·b - a^4·c - a·b^4 + a·c^4 + b^4·c - b·c^4 } {-a^2·b + a^2·c + a·b^2 - a·c^2 - b^2·c + b·c^2 } \\\\ &=& -a^2 - a·b - a·c - b^2 - b·c - c^2 \\ \hline \end{array} }\)

\( a^4(b-c)+b^4(c-a)+c^4(a-b) = (b-c)(c-a)(a-b)(-a^2 - a·b - a·c - b^2 - b·c - c^2) \)

My math program says that you can transform....
\(1/sec\theta * tan\theta = sin\theta\)
Into...
\(tan^2\theta + 1 = sec^2\theta\)
Could someone explain how they made this transition?

\(\begin{array}{|rcll|} \hline \frac{1}{\sec(\theta)} * \tan(\theta) &=& \sin(\theta) \quad & | \quad \cdot \sec(\theta) \\ \tan(\theta) &=& \sin(\theta) \cdot \sec(\theta) \quad & | \quad \text{square both sides} \\ \tan^2(\theta) &=& \sin^2(\theta) \cdot \sec^2(\theta) \quad & | \quad \sin^2(\theta) + \cos^2(\theta) = 1 \text{ or } \sin^2(\theta) = 1-\cos^2(\theta) \\ \tan^2(\theta) &=& [~1-\cos^2(\theta)~] \cdot \sec^2(\theta) \\ \tan^2(\theta) &=& \sec^2(\theta) - \cos^2(\theta) \cdot \sec^2(\theta) \quad & | \quad \sec^2(\theta) = \frac{1}{\cos^2(\theta)} \\ \tan^2(\theta) &=& \sec^2(\theta) - \cos^2(\theta) \cdot \frac{1}{\cos^2(\theta)} \\ \tan^2(\theta) &=& \sec^2(\theta) - \frac{\cos^2(\theta)}{\cos^2(\theta)} \quad & | \quad \frac{\cos^2(\theta)}{\cos^2(\theta)} = 1 \\ \tan^2(\theta) &=& \sec^2(\theta) - 1 \quad & | \quad +1 \\ \mathbf{\tan^2(\theta)+1} &\mathbf{=}& \mathbf{\sec^2(\theta)} \\ \hline \end{array}\)

What is the smallest positive integer that will satisfy the following two modular equations:

N mod 3,331 =1,851 and

N mod 1,851 =1,468?

\(\begin{array}{|rcll|} \hline N &\equiv& 1851 \pmod{3331} \\ N &\equiv& 1468 \pmod{1851} \\ \hline \end{array}\)

Formula:

\(\begin{array}{rcll} N = 1851\cdot 1851\cdot \frac{1}{1851} \pmod{3331} + 1468\cdot 3331 \cdot \frac{1}{3331} \pmod{1851} + 3331\cdot 1851\cdot z \\ \quad z \in Z \\ \end{array}\)

check:

\(\begin{array}{|rcll|} \hline \pmod{3331}: & N = 1851 \cdot \frac{1851}{1851} + 0 +0 = 1851 \ \checkmark \\ \pmod{1851}: & N = 0 + 1468 \cdot \frac{3331}{3331} + 0 = 1468 \ \checkmark \\ \hline \end{array}\)

\(\text{with }\varphi() =\text{ Euler's totient theorem }: \\ 3331 \text{ prime number} \qquad \varphi(p) = p-1 \qquad \varphi(3331) = 3330 \\ 1851 =3\cdot 617\qquad \varphi(1851) =1851\cdot \left(1-\frac13\right)\left(1-\frac{1}{617}\right) = 1232 \)

Modular inverses:

\(\begin{array}{|rcll|} \hline && \frac{1}{1851} \pmod{3331} \\ &\equiv& 1851^{\varphi(3331)-1} \pmod{3331} \quad & | \quad gcd(3331,1851)=1 \\ &\equiv& 1851^{3330-1} \pmod{3331} \\ &\equiv& 1851^{3329} \pmod{3331} \\ && \ldots \\ &\equiv& 835 \pmod{3331} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \frac{1}{3331} \pmod{1851} \\ &\equiv& 3331^{\varphi(1851)-1} \pmod{1851} \quad & | \quad gcd(3331,1851)=1 \\ &\equiv& 3331^{1232-1} \pmod{1851} \\ &\equiv& 3331^{1231} \pmod{1851} \\ && \ldots \\ &\equiv& 1387 \pmod{1851} \\ \hline \end{array}\)

\(\begin{array}{rcll} N &=& 1851\cdot 1851\cdot 835 + 1468 \cdot 3331 \cdot 1387 + 3331\cdot 1851\cdot z \quad & | \quad z \in Z \\ &=& 2860877835 + 6782302396 + 6165681\cdot z \\ &=& \underbrace{9643180231}_{\equiv 55147 \pmod{6165681}} + 6165681\cdot z \\ &=& 55147 + 6165681\cdot z \\ \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{55147} &\mathbf{\equiv}& \mathbf{1851 \pmod{3331}} \\ \mathbf{55147} &\mathbf{\equiv}& \mathbf{1468 \pmod{1851}} \\ \hline \end{array}\)

Compute: 7Chose6

\(\begin{array}{|rcll|} \hline && \mathbf{\binom{7}{6}} \\\\ &=& \binom{7}{7-6} \\\\ &=& \binom{7}{1} \\\\ &=& \dfrac{7}{1} \\\\ &\mathbf{=}&\mathbf{ 7 } \\ \hline \end{array}\)