Solution :
To factor completely
[tex] a^{2} + 3a-28 [/tex]
To factor it completely , first take the product of first and third term, and then break the second term in two parts in such a way that its sum equals the second term and the product equals the product of first and third term.
Product of first and third term is [tex] (a^{2})(-28) = -28 a^{2} [/tex]
Second term is 3a, Re-write 3a as sum of 7a and -4a.
Lets check it Product of 7a and -4a is [tex] -28 a^{2} [/tex] wich is eaqual to product of first and third term and sum of 7a and -4a is 3a which is the second term.
Now factorise, we get
[tex] a^{2}+3a-28\\\\ = a^{2} +7a-4a-28\\\\=a(a+7)-4(a+7)\\\\=(a+7)(a-4) [/tex]
Hence, [tex] (a+7)(a-4) [/tex] is the factor of [tex] a^{2}+3a-28 [/tex].
