Algebra-1

Since -x² + bx - 10 = -(x + m)(x + n) = -x² - (m + n)x - mn.

For x² + bx - 10, here a = -1 and c = -10.

To find two numbers m and n such that mn = 10 > 0, the possible combinations are m = -2, n = -5; or m = 2, n = 5; or m = -10, n = -1; or m = 10, n = 1.

When m = -2, n = -5, b = -(m + n) = -(-2 - 5) = 8;
when m = 2, n = 5<...