1. **State the problem:** We have a rectangular cardboard 30 cm by 20 cm. Squares of side length $x$ cm are cut from each corner, and the sides are folded up to form an open box. The volume of the box is given as 832 cm³. We need to find the dimensions of the box (length, width, height) when the volume is 832 cm³. 2. **Write the formula for volume of the box:** The height of the box is $x$ (the side length of the cut squares). The new length after cutting is $30 - 2x$ (since squares are cut from both ends). The new width after cutting is $20 - 2x$. Volume $V$ is given by: $$V = \text{length} \times \text{width} \times \text{height} = (30 - 2x)(20 - 2x)(x)$$ 3. **Set up the equation with given volume:** $$ (30 - 2x)(20 - 2x)(x) = 832 $$ 4. **Expand the expression:** First expand $(30 - 2x)(20 - 2x)$: $$ (30)(20) - 30(2x) - 2x(20) + (2x)(2x) = 600 - 60x - 40x + 4x^2 = 600 - 100x + 4x^2 $$ So volume equation becomes: $$ (600 - 100x + 4x^2) x = 832 $$ 5. **Distribute $x$:** $$ 600x - 100x^2 + 4x^3 = 832 $$ 6. **Rewrite as a cubic equation:** $$ 4x^3 - 100x^2 + 600x - 832 = 0 $$ 7. **Simplify by dividing entire equation by 4:** $$ \cancel{4}x^3 - \cancel{100}x^2 + \cancel{600}x - \cancel{832} = 0 \Rightarrow x^3 - 25x^2 + 150x - 208 = 0 $$ 8. **Solve the cubic equation for $x$:** We look for reasonable positive roots less than 10 (since $x$ must be less than half the smaller side 20/2=10). Try $x=4$: $$4^3 - 25(4)^2 + 150(4) - 208 = 64 - 400 + 600 - 208 = 56 \neq 0$$ Try $x=3$: $$27 - 225 + 450 - 208 = 44 \neq 0$$ Try $x=2$: $$8 - 100 + 300 - 208 = 0$$ So $x=2$ is a root. 9. **Find other roots by polynomial division:** Divide $x^3 - 25x^2 + 150x - 208$ by $(x-2)$: Using synthetic division: 2 | 1 -25 150 -208 | 2 -46 208 ---------------- 1 -23 104 0 So quotient is $x^2 - 23x + 104$. 10. **Solve quadratic $x^2 - 23x + 104 = 0$:** Use quadratic formula: $$x = \frac{23 \pm \sqrt{23^2 - 4 \times 1 \times 104}}{2} = \frac{23 \pm \sqrt{529 - 416}}{2} = \frac{23 \pm \sqrt{113}}{2}$$ Calculate approximate roots: $$\sqrt{113} \approx 10.63$$ So, $$x_1 = \frac{23 + 10.63}{2} = 16.82$$ $$x_2 = \frac{23 - 10.63}{2} = 6.18$$ 11. **Check feasible values:** $x$ must be less than 10 (half of 20 cm), so $x=16.82$ is invalid. Valid roots: $x=2$ and $x=6.18$. 12. **Calculate box dimensions for each valid $x$:** For $x=2$: Length = $30 - 2(2) = 26$ cm Width = $20 - 2(2) = 16$ cm Height = $2$ cm For $x=6.18$: Length = $30 - 2(6.18) = 30 - 12.36 = 17.64$ cm Width = $20 - 2(6.18) = 20 - 12.36 = 7.64$ cm Height = $6.18$ cm 13. **Verify volume for both:** For $x=2$: $$26 \times 16 \times 2 = 832$$ (matches exactly) For $x=6.18$: $$17.64 \times 7.64 \times 6.18 \approx 832$$ (approximate check) 14. **Final answer:** The box dimensions can be either: - Length = 26 cm, Width = 16 cm, Height = 2 cm - Length = 17.64 cm, Width = 7.64 cm, Height = 6.18 cm Rounded to nearest hundredth: $$\boxed{(26, 16, 2) \text{ or } (17.64, 7.64, 6.18)}$$