Multiplying Polynomials: Practice

Distribute $2x$ and then $-3$ across the trinomial, giving six products.

$6x^3 - 2x^2 + 8x - 9x^2 + 3x - 12$

Combine: $-2x^2 - 9x^2 = -11x^2$ and $8x + 3x = 11x$, so the product is $6x^3 - 11x^2 + 11x - 12$.

Multiply the first two factors, then multiply by the third.

$(x + 2)(x - 1) = x^2 + x - 2$

Now multiply by $(x + 3)$: $x^3 + 3x^2 + x^2 + 3x - 2x - 6 = x^3 + 4x^2 + x - 6$.

The $x^2$ terms come from $2x^2 \cdot (-2)$, $(-3x)(4x)$, and $1 \cdot x^2$.

$-4x^2 - 12x^2 + x^2 = -15x^2$

So the coefficient of $x^2$ is $-15$.

Expanding gives $x^2 + (k + 3)x + 3k$. Match the constant term and solve for $k$.

$3k = 12 \;\Rightarrow\; k = 4$

The middle coefficient confirms it: $k + 3 = 4 + 3 = 7$, matching the $7x$ in the target.

The $x$ term is the sum of the outer and inner products, $-3x$ and $5x$.

$-3x + 5x = 2x$

So the coefficient of $x$ is $2$.

The degree of a product is the sum of the degrees, and each factor has degree $2$.

$2 + 2 + 2 = 6$

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

$-x^3 + 6x^3 = 5x^3$

So the coefficient of $x^3$ is $5$.

Multiply the first two factors first.

$(2x + 1)(3x - 2) = 6x^2 - x - 2$

Now multiply by $(x + 1)$: $6x^3 + 6x^2 - x^2 - x - 2x - 2 = 6x^3 + 5x^2 - 3x - 2$.

Distribute; every interior term cancels with the next in a telescoping chain.

$x^4 + x^3 + x^2 + x - x^3 - x^2 - x - 1 = x^4 - 1$

The $x$ terms come from $(-2x)(-5)$ and $3 \cdot 2x$.

$10x + 6x = 16x$

So the coefficient of $x$ is $16$.

Square with $a = 3x^2$ and $b = 2$; the middle term is $-2ab$.

$(3x^2)^2 - 2(3x^2)(2) + 2^2 = 9x^4 - 12x^2 + 4$

The only way to reach $x^4$ is $x^2 \cdot x^2$, since $x^2$ is the highest power in each factor.

$1 \cdot 1 = 1$

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