Math Problem Statement

how would you factorise this quartic?

How would you factorise this quartic?

Solution

To factorize the quartic x42x33x2+4x+4x^4 - 2x^3 - 3x^2 + 4x + 4, let's follow these steps:


Step 1: Look for Rational Roots

Using the Rational Root Theorem, the possible roots are the factors of the constant term 44 divided by the leading coefficient (which is 11). These factors are: ±1,±2,±4\pm 1, \pm 2, \pm 4

We test these by substituting into the polynomial.

  1. Testing x=1x = 1: f(1)=142(13)3(12)+4(1)+4=123+4+4=4(Not a root)f(1) = 1^4 - 2(1^3) - 3(1^2) + 4(1) + 4 = 1 - 2 - 3 + 4 + 4 = 4 \quad \text{(Not a root)}

  2. Testing x=1x = -1: f(1)=(1)42(1)33(1)2+4(1)+4=1+234+4=0(Root found!)f(-1) = (-1)^4 - 2(-1)^3 - 3(-1)^2 + 4(-1) + 4 = 1 + 2 - 3 - 4 + 4 = 0 \quad \text{(Root found!)}

Thus, x+1x + 1 is a factor.


Step 2: Polynomial Division

Divide x42x33x2+4x+4x^4 - 2x^3 - 3x^2 + 4x + 4 by x+1x + 1 using synthetic or long division.

Long Division Process:

Divide: x42x33x2+4x+4x+1\text{Divide: } \frac{x^4 - 2x^3 - 3x^2 + 4x + 4}{x + 1}

  1. Divide the leading term: x4÷x=x3x^4 \div x = x^3.
  2. Multiply: x3(x+1)=x4+x3x^3(x + 1) = x^4 + x^3.
  3. Subtract: (x42x33x2+4x+4)(x4+x3)=3x33x2+4x+4(x^4 - 2x^3 - 3x^2 + 4x + 4) - (x^4 + x^3) = -3x^3 - 3x^2 + 4x + 4.
  4. Repeat for each term:
    • 3x3÷x=3x2-3x^3 \div x = -3x^2,
    • Multiply: 3x2(x+1)=3x33x2-3x^2(x + 1) = -3x^3 - 3x^2,
    • Subtract: (3x33x2+4x+4)(3x33x2)=4x+4(-3x^3 - 3x^2 + 4x + 4) - (-3x^3 - 3x^2) = 4x + 4,
    • 4x÷x=44x \div x = 4,
    • Multiply: 4(x+1)=4x+44(x + 1) = 4x + 4,
    • Subtract: (4x+4)(4x+4)=0(4x + 4) - (4x + 4) = 0.

Result of division: x42x33x2+4x+4=(x+1)(x33x23x+4)x^4 - 2x^3 - 3x^2 + 4x + 4 = (x + 1)(x^3 - 3x^2 - 3x + 4)


Step 3: Factorize the Cubic x33x23x+4x^3 - 3x^2 - 3x + 4

Repeat the process to factorize the cubic.

  1. Test for roots of x33x23x+4x^3 - 3x^2 - 3x + 4:
    • f(1)=133(12)3(1)+4=133+4=1(Not a root)f(1) = 1^3 - 3(1^2) - 3(1) + 4 = 1 - 3 - 3 + 4 = -1 \quad (\text{Not a root}),
    • f(1)=(1)33(1)23(1)+4=13+3+4=3(Not a root)f(-1) = (-1)^3 - 3(-1)^2 - 3(-1) + 4 = -1 - 3 + 3 + 4 = 3 \quad (\text{Not a root}),
    • f(2)=233(22)3(2)+4=8126+4=6(Not a root)f(2) = 2^3 - 3(2^2) - 3(2) + 4 = 8 - 12 - 6 + 4 = -6 \quad (\text{Not a root}),
    • f(2)=(2)33(2)23(2)+4=812+6+4=10(Not a root)f(-2) = (-2)^3 - 3(-2)^2 - 3(-2) + 4 = -8 - 12 + 6 + 4 = -10 \quad (\text{Not a root}).

None of the integer roots work, suggesting it may not factor further nicely.


Final Answer:

The factorization is: x42x33x2+4x+4=(x+1)(x33x23x+4)x^4 - 2x^3 - 3x^2 + 4x + 4 = (x + 1)(x^3 - 3x^2 - 3x + 4) You could explore further factorization of the cubic using advanced methods (numerical approximations or factoring techniques).


Would you like to go deeper into solving the cubic part or need more clarification?

Related Questions:

  1. How can synthetic division simplify polynomial division?
  2. What is the Rational Root Theorem, and how is it applied?
  3. How can you verify the factorization of a polynomial?
  4. When is it possible to completely factorize a cubic polynomial?
  5. How are complex roots identified in polynomial factorization?

Tip:

Always test integer roots first using the Rational Root Theorem before attempting more advanced factorization techniques!

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Math Problem Analysis

Mathematical Concepts

Algebra
Polynomial Factorization
Rational Root Theorem
Polynomial Long Division

Formulas

Rational Root Theorem
Synthetic Division

Theorems

Fundamental Theorem of Algebra

Suitable Grade Level

Grades 10-12