Practice questions

$=\dfrac{3^{-4}x^{5(-4)}y^{-3(-4)}}{9xy^2}$

$=\dfrac{3^{-4}x^{-20}y^{12}}{9xy^2}$

$=\dfrac{x^{-20}y^{12}}{3^4\cdot 9xy^2}$

$=\dfrac{y^{12-2}}{81\cdot 9x^{1+20}}$

$=\dfrac{y^{10}}{729x^{21}}$

Use the formula, $(a-b)^2=a^2-2ab+b^2$

$=(3x)^2-2\cdot 3x\cdot 7y+(7y)^2$

$=9x^2-42xy+49y^2$

$=\dfrac{(x+5)(x-2)}{x+5}$

$=x-2$

Factor out the greatest common factor:

$=8x^2y^6(8x^7y^3+3)$

Factor by grouping.

$=\underline{m^3+4m^2}\:\:\underline{-6m-24}$

$=m^2(m+4)-6(m+4)$

$=(m+4)(m^2-6)$

$8x^2-6x-9$

This is a quadratic. Multiplying the coefficient of $x$ and the constant term, you get $-72$. Find two numbers whose product is $-72$ and their sum is $-6$. The two numbers are $-12$ and $6$. Write the $x$ term as $-12x+6x$:

$=8x^2-12x+6x-9$

Factor by grouping:

$=\underline{8x^2-12x}+\underline{6x-9}$

$=4x(2x-3)+3(2x-3)$

$=(2x-3)(4x+3)$

You can write the expression as a difference of squares:

$=(4x)^2-9^2$

Use the difference of squares formula: $a^2-b^2=(a+b)(a-b)$

$=(4x+9)(4x-9)$

You can write the expression as a sum of cubes:

$=(2c)^3+5^3$

Use the sum of cubes formula: $a^3+b^3=(a+b)(a^2-ab+b^2)$

$=(2c+5)((2c)^2-2c\cdot 5+5^2)$

$=(2c+5)(4c^2-10c+25)$

Factor by grouping: take the first three terms together as you can write it as a perfect square:

$=\underline{x^2+6x+9}-4y^2$

Factoring the quadratic that is underlined, you get: $(x+3)(x+3)$$=(x+3)^2$.

$=\underline{(x+3)^2}-4y^2$

This can be written as a difference of squares:

$=(x+3)^2-(2y)^2$

Apply the difference of squares formula: $a^2-b^2=(a+b)(a-b)$:

$=(x+3+2y)(x+3-2y)$

Write the equation in standard form of a quadratic equation:

$2k^2-9k+9=0$

Factor:

$2k^2-6k-3k+9=0$

$2k(k-3)-3(k-3)=0$

$(k-3)(2k-3)=0$

By zero-product rule:

$k-3=0$ or $2k-3=0$

Solve for k:

$k=3$ or $k=3/2$

Solution:

$\{3, 3/2\}$

First factor the expression $k^2-4$ in the denominator. This you can write as a difference of squares: $k^2-2^2=(k+2)(k-2)$:

$\dfrac{3}{k+2}-\dfrac{2}{(k+2)(k-2)}=\dfrac{1}{k-2}$

Least common denominator is $(k+2)(k-2)$

Multiply this in each term of the equation:

$\dfrac{3}{k+2}\cdot (k+2)(k-2)-\dfrac{2}{(k+2)(k-2)}\cdot (k+2)(k-2)$

  $=\dfrac{1}{k-2}\cdot (k+2)(k-2)$

Canceling:

$\dfrac{3}{\cancel{k+2}}\cdot (\cancel{k+2})(k-2)-\dfrac{2}{(\cancel{k+2})(\cancel{k-2})}\cdot (\cancel{k+2})(\cancel{k-2})$

  $=\dfrac{1}{\cancel{k-2}}\cdot (k+2)(\cancel{k-2})$

You get,

$3(k-2)-2=k+2$

Now, solve for $k$:

$3k-6-2=k+2$

$2k=10$

$k=5$

Since the original equation is a rational equation, you need to check whether the $k=5$ is the solution by substituting this in the original equation. We can confirm, $5$ is the solution.

Thus, the solution is $k=5$.

You can factor $m^2-49$ as $(m+7)(m-7)$:

$=\dfrac{(m+7)(m-7)}{m+1}\div \dfrac{7-m}{m}$

Replace the multiplication by taking the reciprocal of the rational expression that follows the division sign:

$=\dfrac{(m+7)(m-7)}{m+1}\cdot \dfrac{m}{7-m}$

Write $7-m=-(m-7)$ as this and the $(m-7)$ in the denominator, you can cancel out:

$=\dfrac{(m+7)(m-7)}{m+1}\cdot \dfrac{m}{-(m-7)}$

Canceling $(m-7)$:

$=-\dfrac{m(m+7)}{m+1}$

Factor the expressions in the denominators:

$=\dfrac{5x}{(x+2y)(x-y)}-\dfrac{3x}{(x+6y)(x-y)}$

LCD$=(x+2y)(x-y)(x+6y)$

In each term, divide the $LCD$ by the denominator and multiply that. And put the LCD as the common denominator:

$=\dfrac{5x(x+6y)-3x(x+2y)}{(x+2y)(x-y)(x+6y)}$

Multiply and remove the parentheses in the numerator:

$=\dfrac{5x^2+30xy-3x^2-6xy}{(x+2y)(x-y)(x+6y)}$

$=\dfrac{2x^2+24xy}{(x+2y)(x-y)(x+6y)}$

Factor the numerator:

$=\dfrac{2x(x+12y)}{(x+2y)(x-y)(x+6y)}$