Question

(P+1) (p+3) (P+5) (p+7) +15: factorisation of middle terms​

Answer 1

(p+1)(p+3)(p+5)(p+7) + 15

= [(p+1)(p+7)][(p+3)(p+5) ]+ 15

= (p²+8p+7)(p²+8p+15)+15

Let p²+8p+7 = a ---(1)

= a(a+8)+15

= a²+8a+15

/* splitting the middle term we get */

= a² + 5a + 3a + 15

= a(a+5) + 3(a+5)

= (a+5)(a+3)

= (p²+8p+7+5)(p²+8p+7+3)

= (p²+8p+12)(p²+8p+10)

= ( p² + 6p + 2p + 12)(p²+8p+10)

= [p(p+6)+2(p+6)](p²+8p+10)

= (p+6)(p+2)(p²+8p+10)

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