Solve x^3 - 14x^2 + 23 = 0.

Step-by-step solution

1. 1

   Note what the missing $x$ term tells you. Written in full the cubic is $x^{3}-14x^{2}+0\cdot x+23=0$, so Vieta gives $r_{1}+r_{2}+r_{3}=14,\qquad r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}=0,\qquad r_{1}r_{2}r_{3}=-23.$ The middle relation being exactly zero is a strong constraint and the best check available at the end.

2. 2

   Rule out rational roots. A rational root of a monic integer cubic must divide the constant term $23$, which is prime, so the only candidates are $\pm1$ and $\pm23$. Testing them: $f(1)=10$, $f(-1)=8$, $f(23)=4784$ and $f(-23)=-19550$. None vanish, so the cubic does not factor over the rationals and the roots must be found numerically.

3. 3

   Bracket the three roots. With $f(x)=x^{3}-14x^{2}+23$: $f(-1.3)=-2.857,\quad f(-1.2)=1.112,\quad f(1.3)=1.537,\quad f(1.4)=-1.696,\quad f(13.8)=-15.088,\quad f(13.9)=3.679.$ Three sign changes, so all three roots are real and they lie in $(-1.3,-1.2)$, $(1.3,1.4)$ and $(13.8,13.9)$.

4. 4

   Refine with Newton's method. Using $f'(x)=3x^{2}-28x$ and iterating $x\mapsto x-\dfrac{f(x)}{f'(x)}$ inside each bracket gives $x_{1}=-1.228935648,\qquad x_{2}=1.348309644,\qquad x_{3}=13.880626004.$

5. 5

   Confirm with all three Vieta relations. The sum is $14.000000000$, the pairwise product sum is $-0.000000000$, and the product is $-23.000000000$ - all three match the coefficients. The pairwise sum is the decisive test: rounded values such as $-1.255$, $1.333$ and $13.922$ can look plausible and even keep the sum near $14$, yet they fail here (and indeed $f(13.922)=7.88$, not $0$).

6. 6

   State the answer. $x\approx -1.228936,\qquad x\approx 1.348310,\qquad x\approx 13.880626.$ Two roots sit close to the origin and one far out near $14$, which is what a dominant $-14x^{2}$ against a small constant $23$ should produce.