Question

Сократите дробь:

4c^2 + 7c - 2

1 - 16c^2

Answer 1

$ax^2+bx+c=a(x-x_1)(x-x_2)$
$D=b^2-4ac$
$x_1=frac{-b-sqrt{D}}{2a}$
$x_2=frac{-b+sqrt{D}}{2a}$
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$4c^2+7c-2=0$
$D=7^2-4*4*(-2)=49+32=81=9^2$
$c_1=frac{-7-9}{2*4}=-2$
$c_2=frac{-7+9}{2*4}=frac{1}{4}$
$4c^2+7c-2=4(c-(-2))(c-frac{1}{4})=(c+2)(4c-1)$
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$1-16c^2=1^2-(4c)^2=(1-4c)(1+4c)=-(4c+1)(4c-1)$
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$frac{4c^2+7c-2}{1-16c^2}=frac{(c+2)(4c-1)}{-(4c+1)(4c-1)}=\\-frac{c+2}{4c+1}$