Practice 1 Solutions

Determine if the given expressions will form an identity.

${(a + b)}^{3}$ and $a (a^{2} + a b) + 2 a b (a + b) + b^{2} (a + b)$

Left side: ${(a + b)}^{3}$	Right side: $a (a^{2} + a b) + 2 a b (a + b) + b^{2} (a + b)$
$(a + b) (a + b) (a + b) (a^{2} + 2 a b + b^{2}) (a + b) a^{3} + a^{2} b + 2 a^{2} b + 2 a b^{2} + a b^{2} + b^{3} a^{3} + 3 a^{2} b + 3 a b^{2} + {b^{3}}^{}$	$a^{3} + a^{2} b + 2 a^{2} b + 2 a b^{2} + a b^{2} + b^{3} a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$

Yes, this is an identity.

$- 6 x^{2} - 2 (25 x + 28)$ and $3 x (- 2 x - 14) - 8 (x + 7)$

- 6 x^{2} - 50 x - 56

- 6 x^{2} - 42 x - 8 x - 56 - 6 x^{2} - 50 x - 56

Yes, this is an identity.

$(\frac{1}{2} n - 7) (n + 8)$ and $\frac{1}{2} n (n + 26) - 56 (n + 1)$

\frac{1}{2} n^{2} + 4 n - 7 n - 56 \frac{1}{2} n^{2} - 3 n - 56

\frac{1}{2} n^{2} + 13 n - 56 n - 56 \frac{1}{2} n^{2} - 43 n - 56

No, this is not an identity.

${(3 g)}^{2} + 10 (3 g + 2) + 1$ and ${(3 g)}^{2} + 10 (3 g + 3) - 9$

9 g^{2} + 30 g + 20 + 1 9 g^{2} + 30 g + 21

9 g^{2} + 30 g + 30 - 9 9 g^{2} + 30 g + 21

Yes, this is an identity.

$x (5 x + 8) (5 x - 8)$ and $x {(5 x)}^{2} - 8^{2}$

x (25 x^{2} - 64) 25 x^{2} - 64 x

x (25 x^{2}) - 64 25 x^{3} - 64

No, this is not an identity.

${(2 a b)}^{2} + 4 (5 a b) + {(5 a b)}^{2}$ and $4 a b (a b + 4) + a b (25 a b + 4)$

4 a^{2} b^{2} + 20 a b + 25 a^{2} b^{2} 29 a^{2} b^{2} + 20 a b

4 a^{2} b + 16 a b + 25 a^{2} b^{2} + 4 a b 29 a^{2} b^{2} + 20 a b

Yes, this is an identity.

Find the missing value(s) in the given equation.

Note

Problems 7–12
You can check your work using substitution.

$(R x + 4) (2 x - R) = 6 x^{2} - x - 12$

$\begin{array}{rcl} 2 R x^{2} - R^{2} x + 8 x - 4 R & = & 6 x^{2} - x - 12 \\ 2 R x^{2} & = & 6 x^{2} \\ 2 R & = & 6 \\ R & = & 3 \end{array}$

$(B x^{2} - P x + 8) + (4 x^{2} + 1 x - 15) = 10 x^{2} - 14 x - 7$

\begin{array}{rcl} B x^{2} + 4 x^{2} - P x + 1 x + 8 - 15 & = & 10 x^{2} - 14 x - 7 \\ B + 4 & = & 10 \\ B & = & 6 \end{array}

\begin{array}{rcl} - P + 1 & = & - 14 \\ - P & = & - 15 \\ P & = & 15 \end{array}

$3 x (W x + Q) + 8 (W x + Q) = 12 x^{2} + 35 x + 8$

$\begin{array}{rcl} 3 W x^{2} + 3 Q x + 8 W x + 8 Q & = & 12 x^{2} + 35 x + 8 \\ 3 W x^{2} & = & 12 x^{2} \\ 3 W & = & 12 \\ W & = & 4 \end{array}$

$\begin{array}{rcl} 3 Q x + 8 W x & = & 35 x \\ 3 Q + 8 W & = & 35 \\ 3 Q + 8 (4) & = & 35 \\ 3 Q + 32 & = & 35 \\ 3 Q & = & 3 \\ Q & = & 1 \end{array}$

Note

First solve for W; then solve for Q.

${(A x + 3)}^{2} = 25 x^{2} + 30 x + 9$

$\begin{array}{rcl} A^{2} x^{2} + 6 A x + 9 & = & 25 x^{2} + 30 x + 9 \\ A^{2} x^{2} & = & 25 x^{2} \\ A^{2} & = & 25 \\ A & = & \pm 5 \\ A & = & 5 \end{array}$

Note

Because the middle term is positive, A must be positive. You may also solve the equation 6A = 30 to find that A = 5.

$(5 x^{2} - G x + 1) - (9 x^{2} - M x + 2) = - G x^{2} + 3 x - 1$

$\begin{array}{rcl} - 4 x^{2} - G x + M x - 1 & = & - G x^{2} + 3 x - 1 \\ - 4 x^{2} & = & - G x^{2} \\ G & = & 4 \end{array}$

$\begin{array}{rcl} - G x + M x & = & 3 x \\ - 4 + M & = & 3 \\ M & = & 7 \end{array}$

Note

First solve for G ; then solve for M.

$(R x + K) (R x - K) = 36 x^{2} - 81$ ; Assume R and K are whole numbers.

\begin{array}{rcl} R^{2} x^{2} - K^{2} & = & 36 x^{2} - 81 \\ R^{2} & = & 36 \\ R & = & \pm 6 \\ R & = & 6 \end{array}

\begin{array}{rcl} - K^{2} & = & - 81 \\ K^{2} & = & 81 \\ K & = & \pm 9 \\ K & = & 9 \end{array}