Math, asked by ssdh8272916, 11 months ago

factorise (a-1)(a+2)(a+3)(a+4)-3

Answers

Answered by harendrachoubay
7

(a-1)(a+2)(a+3)(a+4)-3=a^{4}+8a^{3} +17a^{2} -2a-27

Step-by-step explanation:

We have,

(a-1)(a+2)(a+3)(a+4)-3

To expand (a-1)(a+2)(a+3)(a+4)-3=?

(a-1)(a+2)(a+3)(a+4)-3

=[(a-1)(a+2)][(a+3)(a+4)]-3

=(a^{2}+2a-a-2)(a^{2}+4a+3a+12)-3

=(a^{2}+a-2)(a^{2}+7a+12)-3

=a^{4}+7a^{3} +12a^{2} +a^{3} +7a^{2} +12a-2a^{2} -14a-24-3

=a^{4}+8a^{3} +17a^{2} -2a-27

Hence, (a-1)(a+2)(a+3)(a+4)-3=a^{4}+8a^{3} +17a^{2} -2a-27

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