Question

Prove that \[{x^3} - 3{x^2} - 13x + 15\;\] is exactly divisible by \[{x^2} + 2x - 3\], without doing the actual division.

Answer 1

Hint: We'll start by factoring the polynomial \[{x^2} + 2x - 3\] into its two linear parts and then equating them both to zero. When the corresponding values of x are substituted \[{x^3} - 3{x^2} - 13x + 15\;\] , it should equal zero.

Complete step-by-step solution:
Consider the polynomial \[{x^2} + 2x - 3\] and factorize it into linear parts by separating the middle term.
\[ = {x^2} + 2x - 3\]
\[ = {x^2} + 3x - x - 3\]
\[ = (x + 3)(x - 1)\]
We have factors of \[{x^2} + 2x - 3\] as \[(x + 3)\] and \[(x - 1)\]
Now, we have to prove that \[{x^3} - 3{x^2} - 13x + 15\;\] is exactly divisible by \[{x^2} + 2x - 3\] ,
So, the equation \[{x^3} - 3{x^2} - 13x + 15\;\] will be exactly divisible by \[{x^2} + 2x - 3\], if satisfies the factors of \[{x^2} + 2x - 3\].
We have factors of \[{x^2} + 2x - 3\] are \[(x + 3)\] and \[(x - 1)\]
So, x = -3 and x = 1,
We will substitute the value of x in equation \[{x^3} - 3{x^2} - 13x + 15\], if it becomes zero.
It’s completely divisible.
When we put x= -3 in equation \[{x^3} - 3{x^2} - 13x + 15\], we get
\[ = {x^3} - 3{x^2} - 13x + 15\;\]
\[ = {( - 3)^3} - 3{( - 3)^2} - 13 \times ( - 3) + 15\;\]
When, we have simplified the equation, we get
\[ = - 27 - 27 + 39 + 15\;\]
\[ = - 54 + 54\]
\[ = 0\]
Similarly, we will substitute 1 in equation \[{x^3} - 3{x^2} - 13x + 15\], we get
\[ = {x^3} - 3{x^2} - 13x + 15\;\]
\[ = {1^3} - 3 \times {1^2} - 13 \times 1 + 15\]
When, we have simplified the equation, we get
\[ = 16 - 16\]
\[ = 0\]

Note: In this problem, the solutions corresponding to this quadratic equation \[{x^2} + 2x - 3\] should make the remainder zero when these solutions or values of x are substituted in \[{x^3} - 3{x^2} - 13x + 15\], in order to make equation \[{x^3} - 3{x^2} - 13x + 15\;\] to be exactly divisible by \[{x^2} + 2x - 3\]. We are not allowed to use the long division method in this situation.