![(x+a)=[x-(-a)]\\So,\ f(-a)=0](/tpl/images/0460/6801/5d3e2.png)
in f(x)=0, we get![f(x) = 4x^3-13x^2-ax = 0\\f(-a)= [4(-a)^3]-[13(-a)^2]-[a(-a)] = 0\\(-4a^3-13a^2+a^2)=0\\(-4a^3-12a^2)=0\\-4a^3=12a^2\\\frac{a^3}{a^2} =\frac{12}{-4} \\\\a=(-3)](/tpl/images/0460/6801/ccdf4.png)
Need with question 3 just want to ensure my answer is correct before continuing
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History, 18.01.2020 07:31, kaysayfam747
Need with question 3 just want to ensure my answer is correct before continuing
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