1.3.3 Exercises

Simplify the following terms for appropriate numbers

a

b

x

y

z

$\frac{3 x - 6 x y^{2} + 4 x y z}{- 2 x}$ $=$
.

$\frac{3 x - 6 x y^{2} + 4 x y z}{- 2 x} = - \frac{3}{2} + 3 y^{2} - 2 y z .$
$(3 a - 2 b) \cdot (4 a - 6)$ $=$
.

$(3 a - 2 b) \cdot (4 a - 6) = 12 a^{2} - 18 a - 8 a b + 12 b .$
$(2 a + 3 b)^{2} - (3 a - 2 b)^{2}$ $=$
.

$\begin{matrix} (2 a + 3 b)^{2} - (3 a - 2 b)^{2} & = & (4 a^{2} + 12 a b + 9 b^{2}) - (9 a^{2} - 12 a b + 4 b^{2}) \\ = & - 5 a^{2} + 24 a b + 5 b^{2} . \end{matrix}$
$\frac{3 a}{3 a + 6 b} + \frac{2 b}{a + 2 b}$ $=$
.
Adding up the numerators over the least common denominator gives $\frac{3 a}{3 a + 6 b} + \frac{2 b}{a + 2 b} = \frac{3 a}{3 a + 6 b} + \frac{6 b}{3 a + 6 b} = \frac{3 a + 6 b}{3 a + 6 b} = 1$ .

The input must not contain any brackets.

For the following exercises a little more patience is required. Simplify:

$\frac{1}{2} x (4 x + 3 y) + \frac{3}{2} (5 x^{2} - 6 x y)$ $=$
.

$\frac{1}{2} x (4 x + 3 y) + \frac{3}{2} (5 x^{2} - 6 x y) = 2 x^{2} + \frac{3}{2} x y + \frac{15}{2} x^{2} - 9 x y = \frac{19}{2} x^{2} - \frac{15}{2} x y .$
$\frac{18 x^{2} - 48 x y + 32 y^{2}}{12 y - 9 x} \cdot \frac{18 x + 24 y}{9 x^{2} - 16 y^{2}}$ $=$
.
Simplifying the expression gives

$\begin{matrix} \frac{18 x^{2} - 48 x y + 32 y^{2}}{12 y - 9 x} \cdot \frac{18 x + 24 y}{9 x^{2} - 16 y^{2}} & = & 4 \cdot \frac{9 x^{2} - 24 x y + 16 y^{2}}{4 y - 3 x} \cdot \frac{3 x + 4 y}{9 x^{2} - 16 y^{2}} \\ = & 4 \cdot \frac{(3 x - 4 y)^{2}}{4 y - 3 x} \cdot \frac{3 x + 4 y}{9 x^{2} - 16 y^{2}} \\ = & - 4 \cdot \frac{(3 x - 4 y)^{2}}{3 x - 4 y} \cdot \frac{3 x + 4 y}{(3 x + 4 y) (3 x - 4 y)} = - 4 . \end{matrix}$
$(a^{2} + 5 a - 2) (2 a^{2} - 3 a - 9) - (\frac{1}{2} a^{2} + 3 a - 5) (a^{2} - 4 a + 3)$ $=$
.

$(a^{2} + 5 a - 2) (2 a^{2} - 3 a - 9) - (\frac{1}{2} a^{2} + 3 a - 5) (a^{2} - 4 a + 3)$

$\begin{matrix} = & 2 a^{4} + 10 a^{3} - 4 a^{2} - 3 a^{3} - 15 a^{2} + 6 a - 9 a^{2} - 45 a + 18 \\ - (\frac{1}{2} a^{4} + 3 a^{3} - 5 a^{2} - 2 a^{3} - 12 a^{2} + 20 a + \frac{3}{2} a^{2} + 9 a - 15) \\ = & \frac{3}{2} a^{4} + 6 a^{3} - \frac{25}{2} a^{2} - 68 a + 33 . \end{matrix}$

The input must not contain any brackets.

Use a binomial formula to calculate the following squares:

$43^{2}$ $=$
.

$43^{2} = (40 + 3)^{2} = 40^{2} + 2 \cdot 40 \cdot 3 + 3^{2} = 1600 + 240 + 9 = 1849 .$
$97^{2}$ $=$
.

$97^{2} = (90 + 7)^{2} = 90^{2} + 2 \cdot 90 \cdot 7 + 7^{2} = 8100 + 1260 + 49 = 9409 .$
$41^{2} - 38^{2}$ $=$
.

$41^{2} - 38^{2} = (41 + 38) (41 - 38) = 79 \cdot 3 = 237 .$

Apply a binomial formula to expand the product and collect like terms:

$(- 5 x y - 2)^{2}$ $=$
.

$(- 5 x y - 2)^{2} = (- 1)^{2} \cdot (5 x y + 2)^{2} = 25 x^{2} y^{2} + 20 x y + 4 .$
$(- 6 a b + 7 b c) (- 6 a b - 7 b c)$ $=$
.

$(- 6 a b + 7 b c) (- 6 a b - 7 b c) = (- 6 a b)^{2} - (7 b c)^{2} = 36 a^{2} b^{2} - 49 b^{2} c^{2} .$
$(- 6 a b + 7 b c) (- 6 a b + 7 b c)$ $=$
.

$(- 6 a b + 7 b c) (- 6 a b + 7 b c) = (- 6 a b + 7 b c)^{2} = 36 a^{2} b^{2} - 84 a b^{2} c + 49 b^{2} c^{2} .$
$(x^{2} + 3) (- x^{2} - 3)$ $=$
.

$(x^{2} + 3) (- x^{2} - 3) = - (x^{2} + 3)^{2} = - x^{4} - 6 x^{2} - 9 .$

The input must not contain any brackets.

Factorise the following terms as far as possible using one of the binomial formulas:

$4 x^{2} + 12 x y + 9 y^{2}$ $=$
.

$4 x^{2} + 12 x y + 9 y^{2} = (2 x + 3 y)^{2} .$
$64 a^{2} - 96 a + 36$ $=$
.

$64 a^{2} - 96 a + 36 = (8 a - 6)^{2} .$
$25 x^{2} - 16 y^{2} + 15 x + 12 y$ $=$
.

$\begin{matrix} 25 x^{2} - 16 y^{2} + 15 x + 12 y & = & (5 x)^{2} - (4 y)^{2} + 3 \cdot (5 x + 4 y) \\ = & (5 x + 4 y) (5 x - 4 y) + 3 \cdot (5 x + 4 y) \\ = & (5 x + 4 y) (5 x - 4 y + 3) . \end{matrix}$

Factorise the result until it fits into the input field.