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x^2 + x - 6 over x^2 - 7x + 10
simplify fully

Sagot :

RedRicin
[tex]\frac{x^2+x-6}{x^2-7x+10}[/tex]

When you factor a quadratic ax² + bx + c, it becomes (x+m)(x+n), where mn = ac, and m+n = b.

In x²+x-6, we want to find two numbers that multiply to -6 and add to 1.
3 and -2 satisfy both of these criteria, so there are our factors.

[tex]\frac{(x+3)(x-2)}{x^2-7x+10}[/tex]

In x²-7x+10, we want to find two numbers that multiply to 10 and add to -7.
-5 and -2 satisfy both of these criteria, so there are our factors.

[tex]\frac{(x+3)(x-2)}{(x-5)(x-2)}[/tex]

The (x-2) cancels and we're left with

[tex]\boxed{\frac{x+3}{x-5}}[/tex]
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