Question

with actual division prove that x^3 - 3 x^2 - 13 x + 15 is exactly divisible by X ^2 + 2 x - 3​

Answer 1

EXPLANATION :-

$\sf\Large\qquad\quad \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x - 5\\ \begin{array}{cc} \cline{2 - 2}\sf {x}^{2} + 2x - 3 )&\sf \ {x}^{3} - {3x}^{2} - 13x + 15 \\&\sf {x}^{2} +{2x}^{2} - 3x \ \ \downarrow\\ \cline{2-2}& \sf \ \ \ \ -{5x}^{2}- 10x + 15\\ &\sf \ \ \ -{ 5x}^{2}- 10x + 15\\ \cline{2-2} & \sf \ \00 \\ \cline{2-2} \end{array}$

EXPLANATION:-

• First arrange the polynomial according to highest degree i.e x³, x², x , constant term
• Now , take the dividend of first term i.e x³
• Divide the first term of dividend with the first term of divisor i.e x²
• x³/x² = x
• So, the quotient first term is x
• Multiply the quotient first term x with the divisor
• (x²+2x-3)x = x³ + 2x² -3x
• Then subtract them we get -5x² -10x + 15
• Now take the first term that is -5x²
• Divide with the first term of divisor x²
• -5x²/x² = -5
• Now multiply -5 with the divisor x² + 2x-3
• Since the quotient 2nd term will be -5
• (x² + 2x-3)5 = 5x² + 10x - 15
• Now , subtract both The remainder will be 0

It is exactly divisible means remainder must be 0 then If we divide x³-3x²-13x+15 by x²+2x -3 its remainder is 0 So, we can say that it is exactly divisible

Proved!

If you observe the division

• x-5 is the quotient
• 0 is the remainder
• x²+ 2x-3 is the divisor
• x³-3x² -13x +15 is the dividend