Multiplying Polynomials: Additional Practice (Set 1)

Distribute $3x$ and then $2$ across the trinomial.

$6x^3 - 3x^2 + 12x + 4x^2 - 2x + 8$

Combine: $-3x^2 + 4x^2 = x^2$ and $12x - 2x = 10x$, so the product is $6x^3 + x^2 + 10x + 8$.

Square with $a = 2x$ and $b = 5$; the middle term is $-2ab$.

$(2x)^2 - 2(2x)(5) + 5^2 = 4x^2 - 20x + 25$

Expanding gives $2x^2 + (b + 8)x + 4b$. Match the constant term and solve for $b$.

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

The middle coefficient confirms it: $b + 8 = 3 + 8 = 11$, matching the $11x$ in the target.

Multiply the first two factors, then by the third.

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

Now multiply by $(x + 3)$: $x^3 + 3x^2 + 3x^2 + 9x + 2x + 6 = x^3 + 6x^2 + 11x + 6$.

Difference of squares with $a = 5x$ and $b = 2$.

$(5x)^2 - 2^2 = 25x^2 - 4$

The $x$ terms come from $3x \cdot (-5)$ and $(-2)(4x)$.

$-15x - 8x = -23x$

So the coefficient of $x$ is $-23$.

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

$(x^2)^2 + 2(x^2)(3) + 3^2 = x^4 + 6x^2 + 9$

A binomial has $2$ terms and a trinomial has $3$; every term meets every term.

$2 \times 3 = 6$

The expansion is $x^3 + 6x^2 + 11x + 6$. The coefficient of $x^2$ is the sum of the three constants.

$1 + 2 + 3 = 6$

Distribute $2x$ and then $-3$ across the trinomial.

$6x^3 - 4x^2 + 2x - 9x^2 + 6x - 3$

Combine: $-4x^2 - 9x^2 = -13x^2$ and $2x + 6x = 8x$, so the product is $6x^3 - 13x^2 + 8x - 3$.

Group the factors as $((x^2 - 1) + x)((x^2 - 1) - x)$, a difference of squares.

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

Cube means $(2x - 1)(2x - 1)(2x - 1)$. Square first.

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

Now multiply by $(2x - 1)$: $8x^3 - 8x^2 + 2x - 4x^2 + 4x - 1 = 8x^3 - 12x^2 + 6x - 1$.