Question

factorise:p*2+ p-132​

Answer 1

Answer:

$p { }^{2} + p - 132 \\ \\ = p {}^{2} + (12 - 11)p - 132 \\ \\ = p { }^{2} + 12p - 11p - 132 \\ \\ = (p {}^{2} + 12p) - (11p + 132 )\\ \\ = p(p + 12) - 11(p + 12) \\ \\ = (p + 12)(p - 11)......$

HOPE IT HELP UH❤

Answer 2

Answer:

Ans=11,-12

Step-by-step explanation:

p*2+p-132=0

p*2 +(12-11)p -132=0

p*2+12p-11p-132=0

p(p+12)-11(p+12)=0

(p+12)(p-11)=0

either

p+12=0 or p-11=0

p=-12 p=11