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Mathematics

1 answer:

Vilka [71]4 years ago

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Answer:

$\Large \boxed{\sf \bf \ \ \dfrac{x^2-x-6}{x^2-3x+2} \ \ }$

Step-by-step explanation:

Hello, first of all, we will check if we can factorise the polynomials.

$\boxed{x^2+6x+8}\\\\\text{The sum of the zeroes is -6=(-4)+(-2) and the product 8=(-4)*(-2), so}\\\\x^2+6x+8=x^2+2x+4x+8=x(x+2)+4(x+2)=(x+2)(x+4)$

$\boxed{x^2+3x-10}\\\\\text{The sum of the zeroes is -3=(-5)+(+2) and the product -10=(-5)*(+2), so}\\\\x^2+3x-10=x^2+5x-2x-10=x(x+5)-2(x+5)=(x+5)(x-2)$

$\boxed{x^2+2x-15}\\\\\text{The sum of the zeroes is -2=(-5)+(+3) and the product -15=(-5)*(+3), so}\\\\x^2+2x-15=x^2-3x+5x-15=x(x-3)+5(x-3)=(x+5)(x-3)$

$\boxed{x^2+3x-4}\\\\\text{The sum of the zeroes is -3=(-4)+(+1) and the product -4=(-4)*(+1), so}\\\\x^2+3x-4=x^2-x+4x-4=x(x-1)+4(x-1)=(x+4)(x-1)$

Now, let's compute the product.

$\dfrac{x^2+6x+8}{x^2+3x-10}\cdot \dfrac{x^2+2x-15}{x^2+3x-4}\\\\\\=\dfrac{(x+2)(x+4)}{(x+5)(x-2)}\cdot \dfrac{(x+5)(x-3)}{(x+4)(x-1)}\\\\\\\text{We can simplify}\\\\=\dfrac{(x+2)}{(x-2)}\cdot \dfrac{(x-3)}{(x-1)}\\\\\\=\large \boxed{\dfrac{x^2-x-6}{x^2-3x+2}}$