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Basics of Algebra

1. FOIL Method and Distributive Property


The FOIL method is used to multiply two binomial expressions, while the distributive property allows us to multiply a constant with a binomial. Let's demonstrate with an example:

\[ (2x+1)(3x-2) \]

Applying the FOIL method:

\[ \begin{align*} \text{First:} & \quad 2x \cdot 3x = 6x^2 \\ \text{Outer:} & \quad 2x \cdot (-2) = -4x \\ \text{Inner:} & \quad 1 \cdot 3x = 3x \\ \text{Last:} & \quad 1 \cdot (-2) = -2 \\ \end{align*} \]

Combining these results:

\[ 6x^2 - 4x + 3x - 2 \]

Simplifying further:

\[ 6x^2 - x - 2 \]

Now, let's consider the expression multiplied by a constant:

\[ 3(2x+1)(3x-2) \]

Using the distributive property:

\[ \begin{align*} & \quad 3 \cdot 2x + 3 \cdot 1 \cdot 3x - 3 \cdot 2 \\ = & \quad 6x + 9x^2 - 6 \\ \end{align*} \]

Therefore, \( 3(2x+1)(3x-2) \) simplifies to \( 6x + 9x^2 - 6 \), which is 3 times the result of the original expression.

2. Simplifying an Expression


Let's simplify the expression:

\[ (3x^2-1)(2x+3) - (4x-2)(2x+1) \]

To simplify this expression, we will apply the distributive property and perform the necessary multiplications and subtractions.

Using the distributive property:

\[ \begin{align*} & \quad (3x^2)(2x) + (3x^2)(3) - (1)(2x) - (1)(3) - (4x)(2x) - (4x)(1) + (2)(2x) + (2)(1) \\ = & \quad 6x^3 + 9x^2 - 2x - 3 - 8x^2 - 4x + 4x + 2 \\ \end{align*} \]

Combining like terms:

\[ 6x^3 + (9x^2 - 8x^2) + (-2x + 4x) + (-3 + 2) \]

Simplifying further:

\[ 6x^3 + x^2 + 2x - 1 \]

Therefore, the simplified form of \((3x^2-1)(2x+3) - (4x-2)(2x+1)\) is \(6x^3 + x^2 + 2x - 1\).

3. Difference of Squares


The difference of squares concept states that for any two numbers or algebraic expressions, the square of the first term minus the square of the second term can be factored as the product of their sum and difference:

\[a^2 - b^2 = (a + b)(a - b)\]

Let's apply this concept to the example expression: \(16x^2 - 9y^2\).

We can observe that \(16x^2\) is the square of \(4x\) and \(9y^2\) is the square of \(3y\). Therefore, we can rewrite the expression as:

\[16x^2 - 9y^2 = (4x)^2 - (3y)^2\]

Now, applying the difference of squares formula, we can factor it as:

\[(4x)^2 - (3y)^2 = (4x + 3y)(4x - 3y)\]

Thus, the expression \(16x^2 - 9y^2\) can be factored as \((4x + 3y)(4x - 3y)\) using the difference of squares concept.

Factor This:
\[81m^4 - 16n^6\]

To factor this expression, we can apply the difference of squares formula:

\[(9m^2)^2 - (4n^3)^2\]

Applying the difference of squares formula, we have:

\[(9m^2 + 4n^3)(9m^2 - 4n^3)\]

Therefore, the factored form of \(81m^4 - 16n^6\) is \((9m^2 + 4n^3)(9m^2 - 4n^3)\).

3. Matching Coefficients


When we have two equations set equal to each other, such as \(ax^2 + bx + c = 4x^2 + 3x + 7\), we can match the coefficients of the corresponding terms on both sides. In this case, we find that \(a = 4\), \(b = 3\), and \(c = 7\) by comparing the coefficients of \(x^2\), \(x\), and the constant term. This concept allows us to equate the corresponding terms in the equation.
\[ax^2 + bx + c = 4x^2 + 3x + 7 \] \[ \text{ means} \] \[ a = 4\] \[b = 3\] \[c = 7\]

Usually, the phrase "is true for all x" or "is equivalent to" is used in these questions

\[ \text{If the below equation is true for all values of x. Find x.} \] \[ px^2 + pqx + rq = 5x^2 + 10x + 6\]

To solve the equation for all values of \(x\), we can match the coefficients of the corresponding terms on both sides of the equation:

\[p = 5\] \[pq = 10\] \[rq = 6\]

To find the value of \(r\), we can substitute the known values \(p = 5\) and \(q = \frac{10}{p} = 2\) into the equation \(rq = 6\):

\[2r = 6 \Rightarrow r = 3\]

Therefore, the value of \(r\) that satisfies the equation for all values of \(x\) is \(r = 3\).

\[ \text{If the below equation is true for all x, find the value of b.}\] \[ (4x-3)(2x^2+3x+5) = ax^2+bx^2+cx+d \]

To solve this, we can expand the left side of the equation using the distributive property:

\[ (4x-3)(2x^2+3x+5) = 8x^3+12x^2+20x-6x^2-9x-15 \]

Simplifying the expression on the left side, we have:

\[ 8x^3+6x^2+11x-15 = ax^2+bx^2+cx+d \]

Comparing the coefficients of the corresponding terms on both sides of the equation, we can see that:

\[ a = 8, \quad b = 6, \quad c = 11, \quad d = -15 \]

Therefore, the value of \( b \) that satisfies the equation for all values of \( x \) is \( b = 6 \).

\[ \text{If the below equation is true for all x, find the value of d.}\] \[ \frac{{5x+1}}{{2x+3}} - \frac{3}{{4x+d}} = \frac{{20x^2+8x-7}}{{(2x+3)(4x+d)}} \]

To simplify, we find a common denominator and combine the fractions:
\[ \frac{{5x+1}}{{2x+3}} - \frac{3}{{4x+d}} = \frac{{(5x+1)(4x+d) - 3(2x+3)}}{{(2x+3)(4x+d)}} \]
For the numerator, expanding and combining like terms, we have:
\[ 20x^2 + (5d-2)x +d-9 \]
Comparing the constant term with \(20x^2 + 14x - 7\), we get:
\[d-9 = -7 \Rightarrow d =9-7 \] \[ \Rightarrow d = 2\]

Therefore, the value of \(d\) that satisfies the equation is \(2\).

3. Matching Coefficients for No Solutions


\[ \text{If the below equation has no solution, then find \(a\)}\] \[4ax + 5 = 12x + 7\]

Since the equation has no solution, it means the coefficients of \(x\) on both sides must be equal.
\[4ax = 12x \]
We can see that the coefficient of \(x\) on the left side is \(4a\), and on the right side is 12. Therefore, we have:
\[4a = 12 \Rightarrow a=3\]

Hence, if the equation \(4ax + 5 = 12x + 7\) has no solution, then \(a\) is equal to 3.

3. Matching Coefficients for Infinite Solutions


\[ \text{If the below equation has infinte solutions, then find \(a\) and \(b\)}\] \[3ax + 30 = 24x + 5b\]

Since the equation has infinite solutions, it means the coefficients of \(x\) on both sides must be equal. Also the constants must also be equal.
\[3ax = 24x \] \[30 = 5b \]
We can see that the coefficient of \(x\) on the left side is \(3a\), and on the right side is 24. Therefore, we have:
\[3a = 24 \Rightarrow a=8\] \[30 = 5b \Rightarrow b=6\]

Hence, if the equation \(3ax + 30 = 24x + 5b\) has infinte solutions, then \(a\) is equal to 8 and \(b\) is equal to 6.

**problems from pg 72 73