put the varible on the shaded half
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The solutions of | x | = 10 are x = 10 and x = -10 .
All (both) solutions happen to be integers.
There are no other solutions.
Let f(x) = ax³ + 9x² + 4x – 10
g(x) = 0⇒x - 3 = 0
⇒ x = 0 + 3
⇒ x = 3
On dividing f(x) by x - 3, it leaves a remainder 5.
Now keeping, f(3) = 5
⇒a(3)³ + 9(3)² + 4(3) - 10 = 5
⇒ a × 27 + 9 × 9 + 4 × 3 - 10 = 5
⇒ 27a + 81 + 12 - 10 = 5
⇒ 81 + 12 - 10 - 5 = 27a
⇒ 81 + 12 - 15 = 27a
⇒ 93 - 15 = 27a
⇒ 78 = 27a
⇒ a = 27/78
⇒ a = 0.3461
