# Without actual di

Let’s find the roots of the equation (x2 + 2x - 3)

⇒ x2 + 3x – x – 3 = 0

⇒ x(x + 3) – 1(x + 3) = 0

⸫ (x + 3) (x – 1)

Hence, if (x + 3) and (x – 1) satisfies the equation x3 – 3x2 – 13x + 15 = 0, then (x3 – 3x2 – 13x + 15) will be exactly divisible by (x2 + 2x - 3).

For x = -3,

⇒ (-3)3– 3(-3)2– 13(-3) + 15

⇒ -27 – 27 + 39 + 15 = 0

For x = 1,

⇒ 13 – 3(1)2– 13(1) + 15

⇒ 1 – 3 – 13 + 15 = 0

Hence proved.

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