Multiplying Polynomials: Additional Practice (Set 2)

The cross terms cancel, leaving a difference of squares with $a = 3x$ and $b = 2$.

$(3x)^2 - (2)^2 = 9x^2 - 4$

The first two factors are a difference of squares.

$(x - 3)(x + 3) = x^2 - 9$

Now multiply by $(x - 1)$: $x^3 - x^2 - 9x + 9$.

Distribute; the middle terms cancel in pairs.

$x^3 + 2x^2 + 4x - 2x^2 - 4x - 8 = x^3 - 8$

This is the difference-of-cubes pattern.

The only source of $x^3$ is $x^2 \cdot (-x)$, since the constant $2$ cannot reach $x^3$.

$x^2 \cdot (-x) = -x^3$

So the coefficient of $x^3$ is $-1$.

Area is length times width, so expand $(2x + 3)(x + 5)$.

$2x^2 + 10x + 3x + 15 = 2x^2 + 13x + 15$

The degree of a product is the sum of the degrees of the factors.

$3 + 4 + 2 = 9$

The degree of a product is the sum of the degrees; the leading terms multiply and cannot cancel.

$4 + 3 = 7$

Expand each product first.

$(x + 2)(x + 5) = x^2 + 7x + 10, \quad (2x + 1)(x - 3) = 2x^2 - 5x - 3$

Subtract, distributing the minus sign: $x^2 + 7x + 10 - 2x^2 + 5x + 3 = -x^2 + 12x + 13$.

The first two factors give a difference of squares.

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

Now multiply: $(x^2 - 4)(x^2 + 4) = x^4 - 16$, another difference of squares.

The $x^4$ terms come from $x^3 \cdot (-3x)$ and $2x^2 \cdot x^2$.

$-3x^4 + 2x^4 = -x^4$

So the coefficient of $x^4$ is $-1$.

The coefficient of $x$ in the expansion is $a + 6$. Set it to zero.

$a + 6 = 0 \;\Rightarrow\; a = -6$

The cross terms cancel, leaving a difference of squares.

$(4x)^2 - (1)^2 = 16x^2 - 1$