Question

(5a+b)²+(5a+b)(a+2b)-20(a+2b)²​

Answer 1

» Question :

$\sf{(5a + b)^{2} + (5a + b)(a + 2b) - 20(a + 2b)^{2}}$

» To Find :

The factorisation of the Equation.

» We Know :

• $\sf{(a + b)^{2} = a^{2} + 2ab + b^{2}}$

» Solution :

$\purple{\sf{(5a + b)^{2} + (5a + b)(a + 2b) - 20(a + 2b)^{2}}}$

Using the identity ,

$\implies \sf{(a + b)^{2} = a^{2} + 2ab + b^{2}}$

$\implies \sf{(5a)^{2} + 2 \times 5a \times b + (b)^{2} + (5a + b)(a + 2b) - 20(a + 2b)^{2}}$

$\implies \sf{25a^{2} + 10ab + b^{2} + (5a + b)(a + 2b) - 20(a + 2b)^{2}}$

$\implies \sf{25a^{2} + 10ab + b^{2} + 5a(a + 2b) + b(a + 2b) - 20(a + 2b)^{2}}$

$\implies \sf{25a^{2} + 10ab + b^{2} + 5a^{2} + 10ab + ab + 2b^{2} - 20(a + 2b)^{2}}$

$\implies \sf{25a^{2} + 10ab + b^{2} + 5a^{2} + 10ab + ab + 2b^{2} - 20\left((a)^{2} + 2 \times a \times 2b + (2b)^{2}\right)}$

$\implies \sf{25a^{2} + 10ab + b^{2} + 5a^{2} + 10ab + ab + 2b^{2} - 20\left(a^{2} + 4ab + 4b^{2}\right)}$

$\implies \sf{25a^{2} + 10ab + b^{2} + 5a^{2} + 10ab + ab + 2b^{2} - 20\left(a^{2} + 4ab + 4b^{2}\right)}$

$\implies \sf{25a^{2} + 10ab + b^{2} + 5a^{2} + 10ab + ab + 2b^{2} - \left(20 \times a^{2} + 20 \times 4ab + 20 \times 4b^{2}\right)}$

$\implies \sf{25a^{2} + 10ab + b^{2} + 5a^{2} + 10ab + ab + 2b^{2} - \left(20a^{2} + 80ab + 80b^{2}\right)}$

$\implies \sf{25a^{2} + 10ab + b^{2} + 5a^{2} + 10ab + ab + 2b^{2} - 20a^{2} - 80ab - 80b^{2}}$

$\implies \sf{\big(25a^{2} + 5a^{2} + (-20a^{2})\big) + \big(10ab + 10ab + ab + (-80ab)\big) + \big(b^{2} + 2b^{2} + (-80b^{2})\big)}$

$\implies \sf{\big(30a^{2} - 20a^{2})\big) + \big(21ab - 80ab)\big) + \big(3b^{2} - 80b^{2}\big)}$

$\sf{\implies 10a^{2} - 59ab - 77b^{2}}$

$\therefore \purple{\sf{10a^{2} - 59ab - 77b^{2}}}$

Answer 2

see ur full answer in attachment mate