Question

(a2+a)2+4 (a2+a)-12 factorize ​

Answer 1

Answer:

(a²-a)²-8(a²-a)+12

(a²-a)²=(a²)²+a²-2a³=a⁴+a²-2a³

a⁴+a²-2a³-8a²+8a+12=

a⁴-2a³-7a²+8a+12

To factorise this biquadratic equation take out the Factors of 12

P(a)=a⁴-2a³-7a²+8a+12

By hidden trail we find out that 2 satisfies P(a)

Thus by remainder theorm a-2 is a factor

Divide P(a) by (a-2)

The result is a³-7a-6

F(x)=a³-7a-6

By hidden trial we find that 3 satisfies F(x)

Thus by remainder theorm a-3 is a factor of F(x)

Divide F(x) by (a-3)

This will result in a²+3a+2=

(a²+a) +(2a+2)

=[a(a+1) ]+[2(a+1) ]

=(a+2) (a+1)

Thus a⁴-2a³-7a²+8a+12=

(a+2) (a+1) (a-2) (a-3)

Answer:(a²-a)²-8(a²-a)+12= (a+2) (a+1) (a-2) (a-3)

Step-by-step explanation:

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