Problem of the week #40 - December 31st, 2012

[Jameson]

Using no calculator to find the roots, find all 4 zeros of $2x^4+11x^3-24x^2-19x+14$.

Hint: You can use the rational root theorem and the quadratic equation.

Let the product of these 4 roots be A.

What is \(\displaystyle \left(A+14 \right) \left(\frac{671}{7} \right)\)?
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[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) Sudharaka

Honorable mention goes to:

1) veronica1999 (didn't show any work )
2) soroban (small error in last line)

Solution (from MarkFL):

Spoiler

$\displaystyle f(x)=2x^4+11x^3+-24x^2-19x+14=0$

Using the rational roots theorem, we know the rational roots will come from the list:

$\displaystyle \pm\left(1,2,7,14,\frac{1}{2},\frac{7}{2} \right)$

$\displaystyle f(1)=-16$

$\displaystyle f(-1)=0$

$\displaystyle f(2)=0$

$\displaystyle f(-2)=-100$

$\displaystyle f(7)=7280$

$\displaystyle f(-7)=0$

$\displaystyle f(14)=102060$

$\displaystyle f(-14)=42224$

$\displaystyle f\left(\frac{1}{2} \right)=0$

Hence, the 4 roots are: $\displaystyle x = -7,\, -1,\, \frac{1}{2},\, 2$

The product of the 4 roots, as given by Vieta is:

$\displaystyle A=(-1)^4\frac{14}{2}=7$

Hence:

$\displaystyle \left(A+14 \right) \left(\frac{671}{7} \right)=3\cdot671=2013$

Another way to find the roots (from soroban):

Spoiler

Using the Rational Roots Theorem,
. . we find that $x = -1$ is a zero.
$f(x)\:=\: (x+1)(2x^3+9x^2-33x + 14)$

We find that $x = 2$ is a zero of the cubic.
$f(x) \:=\: (x+1)(x-2)(2x^2+13x - 7)$

Finally:
$f(x) \:=\: (x+1)(x-2)(x+7)(2x-1)$

The zeros are: .$x \;=\;-1,\,2,\,-7,\,\frac{1}{2}$