Question

without actual division show that (x³-3x²-13x+15) is exactly divisible by (x²+2x-3)
please its urgent!!!!!!!!!!!!!

Answer 1

Hola Mate !!

Here's your answer :-

${x}^{2} + 2x - 3 \\ = {x}^{2} + 3x - x - 3 \\ = x(x + 3) - 1(x + 3) \\ = (x + 3)(x - 1)$

$x + 3 = 0 \\ = > x = - 3$

$( { - 3})^{3} - 3 \times( { - 3})^{2} - 13 \times - 3 + 15 \\ = > - 27 - 3 \times 9 + 39 + 15 \\ = > - 27 - 27 + 54 \\ = > - 54 + 54 \\ = > 0$

$x - 1 = 0 \\ = > x = 1$

${1}^{3} - 3 \times {1}^{2} - 13 \times 1 + 15 \\ = > 1 - 3 - 13 + 15 \\ = > - 16 + 16 \\ = > 0$

$as \: x + 3 \: and \: x - 1 \: is \: divisible \: by \: {x}^{3} - 3 {x}^{2} - 13x + 15 \: \\ so \: {x}^{2} + 2x - 3 \: is \: also \: divisible$

Hope it helps

Mark me as brainliest if it's correct

Answer 2

Answer:

Hope it will help you....

Step-by-step explanation: