NJSLA Algebra I Arithmetic with Polynomials and Rational Expressions. Practice it free.

Multiply the coefficients: $7 \cdot 1 \cdot 2 = 14\textsf{.}$ Add the exponents on $p$ and on $q$: $p^{1+1} q^{1+1} = p^{2} q^{2}\textsf{.}$ So the product is $14p^{2}q^{2}\textsf{.}$

Distribute the minus sign: $(9x^2 - 4x) - 3x^2 + 4x - 5\textsf{.}$ Combine $x^2$ terms: $9x^2 - 3x^2 = 6x^2\textsf{.}$ The $x$ terms cancel: $-4x + 4x = 0\textsf{.}$ The constant is $-5\textsf{.}$ Result: $6x^2 - 5\textsf{.}$

Expand: $(x - 1)(x - 6) = x^2 - 6x - x + 6 = x^2 - 7x + 6$.

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

Factor: $x^2-6x+9=(x-3)^2$. Set the factor equal to zero: $(x-3)^2=0$, so $(x-3)=0 \Rightarrow x=3$. The zero $3$ has multiplicity $2$, so the zeros are $3$ and $3$.

Multiply the coefficients: $5 \cdot 3 = 15\textsf{.}$ Add the exponents on $x$: $x^{1+1} = x^{2}\textsf{.}$ So the product is $15x^{2}\textsf{.}$

Distribute the minus sign: $(7x^2 + 2x - 3) - 5x^2 + 2x - 1\textsf{.}$ Combine $x^2$ terms: $7x^2 - 5x^2 = 2x^2\textsf{.}$ Combine $x$ terms: $2x + 2x = 4x\textsf{.}$ Combine constants: $-3 - 1 = -4\textsf{.}$ Result: $2x^2 + 4x - 4\textsf{.}$

Expand: $(x + 3)(x + 3) = x^2 + 6x + 9$. Then subtract $9$: $x^2 + 6x + 9 - 9 = x^2 + 6x$.