Question

expand(p/4–q/2+1)² the following​

Answer 1

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

To Expand:

• $\left(\dfrac{p}{4} - \dfrac{q}{2} + 1\right)^2$

Solution:

$\implies \left(\dfrac{p}{4} - \dfrac{q}{2} + 1\right)^2 \\\\ \implies \left(\dfrac{p}{4} + \dfrac{(-q)}{2} + 1\right)^2 \\\\ (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \\\\ \implies \left(\dfrac{p}{4}\right)^2 + \left(\dfrac{-q}{2}\right)^2 + 1^2 + 2\left(\dfrac{p}{4}*\dfrac{-q}{2} + \dfrac{-q}{2}*1 + 1*\dfrac{p}{4} \right) \\\\ \implies \dfrac{p^2}{16} + \dfrac{q^2}{4} + 1 + 2\left(\dfrac{(-pq)}{8} + \dfrac{(-q)}{2} + \dfrac{p}{4}\right) \\\\ \implies \dfrac{p^2}{16} + \dfrac{q^2}{4} + 1 - \dfrac{pq}{4} - q + \dfrac{p}{2} \\\\ \bold {\implies \dfrac{p^2}{16} + \dfrac{q^2}{4} + \dfrac{p}{2} - q - \dfrac{pq}{4} + 1}$

Formula Used:

• $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

Learn More:

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\bf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\bf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\bf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\bf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\9)\bf\: A^{3} - B^{3} = (A-B)(A^{2} + AB + B^{2})\\\\ \end{minipage}}$