Question

x^2+11x+30
×^2+6x-72

Answer 1

$\huge\mathbb{SOLUTION:-}$

●we have to factorise the Given expression

$\sf\underline{\pink{\:\:\:\: Solution:-\:\:\:\:\:}}$

$\sf 1)\:\: x{}^{2}+11x+30\\ \\ \sf:\implies x{}^{2}+(6+5)x+30\\ \\ \sf:\implies x{}^{2}+6x+5x+30\\ \\ \sf:\implies x(x+6)+5(x+6)\\ \\ \sf:\implies{\blue{ (x+6)(x+5)}}$

• Now the 2nd question :-

$\sf 2)\:\: x{}^{2}+6x-72\\ \\ \sf:\implies x{}^{2}+(12-6)x-72\\ \\ \sf:\implies x{}^{2}+12x-6x-72\\ \\ \sf:\implies x(x+12)-6(x+12)\\ \\ \sf:\implies{\purple{ (x+12)(x-6)}}$

Answer 2

Answer:

$\bold\red{Solution↓}$

Factorize:-

• $\sf\green {{x}^{2} + 11x + 30 }$
• $\sf\purple {{x}^{2} +6x - 72}$

$\rule{300}{1}$

$\sf a){x}^{2} + 11x + 30$

→ $\sf {x}^{2} + 5x + 6x + 30$

→ $\sf x(x+5) + 6(x+5)$

→ ${\boxed{\sf{(x+6)(x+5)}}}$

$\rule{300}{1}$

$\sf b) {x}^{2} + 6x - 72$

→ $\sf {x}^{2} + 12x - 6x - 72$

→ $\sf x(x+12)-6(x+12)$

→ ${\boxed{\sf{(x-6)(x+12)}}}$