Sum of 3a^{2} – 2a + 5 and a^{2} – 5a – 7
= 3a^{2} – 2a + 5 + a^{2} – 5a – 7
= 3a^{2} + a^{2} – 2a – 5a + 5 – 7
= 4a^{2} – 7a – 2
And sum of 5a^{2} – 9a + 3 and 2a – a^{2 }- 1
= 5a^{2} – 9a + 3 + 2a – a^{2} – 1
= 5a^{2} – a^{2} – 9a + 2a + 3 – 1
= 4a^{2} – 7a + 2
Now, (4a^{2} – 7a + 2) – (4a^{2} – 7a – 2)
= 4a^{2} – 7a + 2 – 4a^{2} + 7a + 2
= 4a^{2} – 4a^{2} – 7a + 7a + 2 + 2
= 0 + 0 + 4 = 4