Use the factor and remainder theorem to factorise \( h(x)=3 x^{3}-9 x^{2}+12 \) completely.

We want to factorise \[ h(x)=3x^3-9x^2+12 \] completely using the factor and remainder theorems. Follow the steps below: --- ### Step 1. Factor Out the Greatest Common Factor Notice that every term of \( h(x) \) is divisible by \( 3 \). Factor \( 3 \) from the polynomial: \[ h(x) = 3\left(x^3 - 3x^2 + 4\right) \] --- ### Step 2. Find a Root Using the Factor Theorem Now, consider the cubic inside the parentheses: \[ p(x) = x^3 - 3x^2 + 4 \] According to the **factor theorem**, if \( p(c) = 0 \), then \((x-c)\) is a factor of \( p(x) \). We test some possible rational roots. Typical candidates (based on the factors of the constant term) include \( \pm 1, \pm 2, \pm 4 \). - **Test \( x = 1 \):** \[ p(1) = 1^3 - 3(1)^2 + 4 = 1 - 3 + 4 = 2 \quad (\neq 0) \] - **Test \( x = -1 \):** \[ p(-1) = (-1)^3 - 3(-1)^2 + 4 = -1 - 3 + 4 = 0 \] Since \( p(-1) = 0 \), the factor theorem tells us that \((x + 1)\) is a factor of \( p(x) \). --- ### Step 3. Divide to Find the Remaining Factor Now, we divide \[ x^3 - 3x^2 + 4 \] by \( (x+1) \) to obtain the quadratic factor. You can use polynomial long division or synthetic division. Here is a brief description of the division: **Division Setup:** Divide \( x^3 - 3x^2 + 0x + 4 \) by \( (x+1) \). 1. **First term:** Divide \( x^3 \) by \( x \) to get \( x^2 \). Multiply \( (x+1) \) by \( x^2 \) to obtain \( x^3 + x^2 \). 2. **Subtract:** \[ (x^3 - 3x^2) - (x^3 + x^2) = -4x^2 \] 3. **Bring Down:** Bring down the \( 0x \) to get \( -4x^2 + 0x \). 4. **Next term:** Divide \( -4x^2 \) by \( x \) to get \( -4x \). Multiply \( (x+1) \) by \( -4x \) to obtain \( -4x^2 - 4x \). 5. **Subtract:** \[ (-4x^2 + 0x) - (-4x^2 - 4x) = 4x \] 6. **Bring Down:** Bring down the \( +4 \) to get \( 4x + 4 \). 7. **Final term:** Divide \( 4x \) by \( x \) to get \( 4 \). Multiply \( (x+1) \) by \( 4 \) to obtain \( 4x+4 \). 8. **Subtract:** \[ (4x+4) - (4x+4) = 0 \] The quotient is \[ x^2 - 4x + 4. \] So, we have: \[ x^3 - 3x^2 + 4 = (x+1)(x^2-4x+4) \] --- ### Step 4. Factorise the Quadratic The quadratic \[ x^2-4x+4 \] is a perfect square because: \[ x^2-4x+4 = (x-2)^2. \] --- ### Final Factorisation Substituting back, we obtain: \[ h(x)=3\left(x^3-3x^2+4\right)=3(x+1)(x-2)^2. \] --- ### Final Answer \[ \boxed{3(x+1)(x-2)^2} \]