How do you find all the zeros of f(x)=x^4+6x^3-3x^2+24x-28?

Jul 27, 2016

Answer:

Zeros: $1 , - 7 , \pm 2 i$

Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We use this later with $a = x$ and $b = 2 i$.

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$f \left(x\right) = {x}^{4} + 6 {x}^{3} - 3 {x}^{2} + 24 x - 28$

First note that the sum of the coefficients is $0$. That is:

$1 + 6 - 3 + 24 - 28 = 0$

Hence $f \left(1\right) = 0$, $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{4} + 6 {x}^{3} - 3 {x}^{2} + 24 x - 28 = \left(x - 1\right) \left({x}^{3} + 7 {x}^{2} + 4 x + 28\right)$

Note that the ratio between the first and second terms of the remaining cubic polynomial is the same as that between its third and fourth terms, so it can be factored by grouping:

${x}^{3} + 7 {x}^{2} + 4 x + 28$

$= \left({x}^{3} + 7 {x}^{2}\right) + \left(4 x + 28\right)$

$= {x}^{2} \left(x + 7\right) + 4 \left(x + 7\right)$

$= \left({x}^{2} + 4\right) \left(x + 7\right)$

Hence $x = - 7$ is another zero of $f \left(x\right)$

The remaining quadratic is positive for all Real values of $x$, but we can factor it using the difference of squares identity with $a = x$ and $b = 2 i$ ...

${x}^{2} + 4 = {x}^{2} - {\left(2 i\right)}^{2} = \left(x - 2 i\right) \left(x + 2 i\right)$

So the last two zeros of $f \left(x\right)$ are $x = \pm 2 i$