Find the following products:

Solution:

(i) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.

$(x+4)(x+7)$

$=x^{2}+(4+7) x+4 \times 7$

$=x^{2}+11 x+28$

(ii) Here, we will use the identity $(x-a)(x+b)=x^{2}+(b-a) x-a b$.

$(x-11)(x+4)$

$=x^{2}+(4-11) x-11 \times 4$

$=x^{2}-7 x-44$

(iii) Here, we will use the identity $(x+a)(x-b)=x^{2}+(a-b) x-a b$.

$(x+7)(x-5)$

$=x^{2}+(7-5) x-7 \times 5$

$=x^{2}+2 x-35$

(iv) Here, we will use the identity $(x-a)(x-b)=x^{2}-(a+b) x+a b$.

$(x-3)(x-2)$

$=x^{2}-(3+2) x+3 \times 2$

$=x^{2}-5 x+6$

(v) Here, we will use the identity $(x-a)(x-b)=x^{2}-(a+b) x+a b$.

$\left(y^{2}-4\right)\left(y^{2}-3\right)$

$=\left(y^{2}\right)^{2}-(4+3)\left(y^{2}\right)+4 \times 3$

$=y^{4}-7 y^{2}+12$

(vi) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.

$\left(x+\frac{4}{3}\right)\left(x+\frac{3}{4}\right)$

$=x^{2}+\left(\frac{4}{3}+\frac{3}{4}\right) x+\frac{4}{3} \times \frac{3}{4}$

$=x^{2}+\frac{25}{12} x+1$

(vii) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.

$(3 x+5)(3 x+11)$

$=(3 x)^{2}+(5+11)(3 x)+5 \times 11$

$=9 x^{2}+48 x+55$

(viii) Here, we will use the identity $(x-a)(x+b)=x^{2}+(b-a) x-a b$.

$\left(2 x^{2}-3\right)\left(2 x^{2}+5\right)$

$=\left(2 x^{2}\right)^{2}+(5-3)\left(2 x^{2}\right)-3 \times 5$

$=4 x^{4}+4 x^{2}-15$

(ix) Here, we will use the identity $(x+a)(x-b)=x^{2}+(a-b) x-a b$.

$\left(z^{2}+2\right)\left(z^{2}-3\right)$

$=\left(z^{2}\right)^{2}+(2-3)\left(z^{2}\right)-2 \times 3$

$=z^{4}-z^{2}-6$

(x) Here, we will use the identity $(x-a)(x-b)=x^{2}-(a+b) x+a b$.

$(3 x-4 y)(2 x-4 y)$

$=(4 y-3 x)(4 y-2 x) \quad$ (Taking common $-1$ from both parentheses)

$=(4 y)^{2}-(3 x+2 x)(4 y)+3 x \times 2 x$

$=16 y^{2}-(12 x y+8 x y)+6 x^{2}$

$=16 y^{2}-20 x y+6 x^{2}$

(xi) Here, we will use the identity $(x-a)(x-b)=x^{2}-(a+b) x+a b$.

$\left(3 x^{2}-4 x y\right)\left(3 x^{2}-3 x y\right)$

$=\left(3 x^{2}\right)^{2}-(4 x y+3 x y)\left(3 x^{2}\right)+4 x y \times 3 x y$

$=9 x^{4}-\left(12 x^{3} y+9 x^{3} y\right)+12 x^{2} y^{2}$

$=9 x^{4}-21 x^{3} y+12 x^{2} y^{2}$

(xii) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.

$\left(x+\frac{1}{5}\right)(x+5)$

$=x^{2}+\left(\frac{1}{5}+5\right) x+\frac{1}{5} \times 5$

$=x^{2}+\frac{26}{5} x+1$

(xiii) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.

$\left(z+\frac{3}{4}\right)\left(z+\frac{4}{3}\right)$

$=z^{2}+\left(\frac{3}{4}+\frac{4}{3}\right) x+\frac{3}{4} \times \frac{4}{3}$

$=z^{2}+\frac{25}{12} z+1$

(xiv) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.

$\left(x^{2}+4\right)\left(x^{2}+9\right)$

$=\left(x^{2}\right)^{2}+(4+9)\left(x^{2}\right)+4 \times 9$

$=x^{4}+13 x^{2}+36$

(xv) Here, we will use the identity $(x+a)(x+b)=x^{2}+(a+b) x+a b$.

$\left(y^{2}+12\right)\left(y^{2}+6\right)$

$=\left(y^{2}\right)^{2}+(12+6)\left(y^{2}\right)+12 \times 6$

$=y^{4}+18 y^{2}+72$

(xvi) Here, we will use the identity $(x+a)(x-b)=x^{2}+(a-b) x-a b$.

$\left(y^{2}+\frac{5}{7}\right)\left(y^{2}-\frac{14}{5}\right)$

$=\left(y^{2}\right)^{2}+\left(\frac{5}{7}-\frac{14}{5}\right)\left(y^{2}\right)-\frac{5}{7} \times \frac{14}{5}$

$=y^{4}-\frac{73}{35} y^{2}-2$

(xvii) Here, we will use the identity $(x+a)(x-b)=x^{2}+(a-b) x-a b$.

$\left(p^{2}+16\right)\left(p^{2}-\frac{1}{4}\right)$

$=\left(p^{2}\right)^{2}+\left(16-\frac{1}{4}\right)\left(p^{2}\right)-16 \times \frac{1}{4}$

$=p^{4}+\frac{63}{4} p^{2}-4$