Hello,
\(if \ a
eq -3, a
eq 1, a
eq -5\\
\dfrac{2a^2+5a-3}{a^2+8a+15} * \dfrac{a^2+4a-5}{3a^2-a-2} \\
= \dfrac{(a+3)(2a-1)}{(a+5)(a+3)} * \dfrac{(a+5)(a-1)}{(a-1)(3a+2)} \\
= \dfrac{2a-1}{3a+2}\)
explains:
1)
\(a^2+4a-5=a^2+5a-a-5\\
=a(a+5)-(a+5)\\
=(a+5)(a-1)\)
2)
\(a^2+8a+15=a^2+5a+3a+15\\
=a(a+5)+3(a+5)\\
=(a+5)(a+3)\)
3)
\(a^2+4a-5=a^2+5a-a-5\\
=a(a+5)-(a+5)\\
=(a+5)(a-1)\)
4)
\(3a^2-a-2=3a^2-3a+2a-2\\
=3a(a-1)+2(a-1)\\
=(a-1)(3a+2)\)
