Find the zeros of the function h(x)=x^2-5x-50 by factoring

1 answer:

5 0

$h(x)=x^2-5x-50=x^2-5x-25-25=x^2-25-5x-25\\\\=(x^2-25)-(5x+25)=(x^2-5^2)-5(x+5)\\\\=(x-5)\underline{(x+5)}-5\underline{(x+5)}=(x+5)(x-5-5)=(x+5)(x-10)\\\\Answer:Zeros\ are\ x=-5\ and\ x=10.$

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