Polynomial Factoring worksheet

1. Factored form: −(4x^4 + 19x − 4)

1. Write the polynomial in standard form, highest power first

   $\text{p(x) = −4x^4 − 19x + 4}$

2. Take out the constant factor −1, leaving 4x^4 + 19x − 4

   $\text{−1(4x^4 + 19x − 4)}$

3. By the rational root theorem, any rational root is ±(a factor of 4) ÷ (a factor of 4)

   $\text{1/4, −1/4, 1/2, −1/2, 1, −1, 2, −2, 4, −4}$

4. 4x^4 + 19x − 4 has no rational roots and no split into two quadratics with integer coefficients, so it is irreducible over the rationals

   $\text{4x^4 + 19x − 4}$

5. Approximate roots of 4x^4 + 19x − 4, found numerically (Durand–Kerner iteration)

   $\text{x ≈ −1.746, x ≈ 0.2101, x ≈ 0.7679 ± 1.4615i}$

6. Put the factors together. Multiplying them back out gives the original polynomial

   $\text{−4x^4 − 19x + 4 = −(4x^4 + 19x − 4)}$

2. Factored form: 3x^5 − 3x^4 − 11x + 6

1. Write the polynomial in standard form, highest power first

   $\text{p(x) = 3x^5 − 3x^4 − 11x + 6}$

2. By the rational root theorem, any rational root is ±(a factor of 6) ÷ (a factor of 3)

   $\text{1/3, −1/3, 2/3, −2/3, 1, −1, 2, −2, 3, −3, 6, −6}$

3. 3x^5 − 3x^4 − 11x + 6 has no rational roots. Factors of degree 2 or more aren't searched for at degree 5, so it may factor further

   $\text{3x^5 − 3x^4 − 11x + 6}$

4. Approximate roots of 3x^5 − 3x^4 − 11x + 6, found numerically (Durand–Kerner iteration)

   $\text{x ≈ −1.31, x ≈ 0.0901 ± 1.3345i, x ≈ 0.5351, x ≈ 1.5948}$

5. Put together the factors found. The last factor may break down further

   $\text{3x^5 − 3x^4 − 11x + 6 = 3x^5 − 3x^4 − 11x + 6}$

3. Factored form: x^3 − 13x + 1

1. Write the polynomial in standard form, highest power first

   $\text{p(x) = x^3 − 13x + 1}$

2. By the rational root theorem, any rational root is ±(a factor of 1) ÷ (a factor of 1)

   $\text{1, −1}$

3. x^3 − 13x + 1 has no rational roots, and a cubic without a rational root can't be factored over the rationals

   $\text{x^3 − 13x + 1}$

4. Approximate roots of x^3 − 13x + 1, found numerically (Durand–Kerner iteration)

   $\text{x ≈ −3.6434, x ≈ 0.077, x ≈ 3.5665}$

5. Put the factors together. Multiplying them back out gives the original polynomial

   $\text{x^3 − 13x + 1 = x^3 − 13x + 1}$

4. Factored form: 2x^5 − 3x + 6

1. Write the polynomial in standard form, highest power first

   $\text{p(x) = 2x^5 − 3x + 6}$

2. By the rational root theorem, any rational root is ±(a factor of 6) ÷ (a factor of 2)

   $\text{1/2, −1/2, 1, −1, 3/2, −3/2, 2, −2, 3, −3, 6, −6}$

3. 2x^5 − 3x + 6 has no rational roots. Factors of degree 2 or more aren't searched for at degree 5, so it may factor further

   $\text{2x^5 − 3x + 6}$

4. Approximate roots of 2x^5 − 3x + 6, found numerically (Durand–Kerner iteration)

   $\text{x ≈ −1.3839, x ≈ −0.2757 ± 1.285i, x ≈ 0.9676 ± 0.5646i}$

5. Put together the factors found. The last factor may break down further

   $\text{2x^5 − 3x + 6 = 2x^5 − 3x + 6}$

5. Factored form: 3x^5 − 17x + 7

1. Write the polynomial in standard form, highest power first

   $\text{p(x) = 3x^5 − 17x + 7}$

2. By the rational root theorem, any rational root is ±(a factor of 7) ÷ (a factor of 3)

   $\text{1/3, −1/3, 1, −1, 7/3, −7/3, 7, −7}$

3. 3x^5 − 17x + 7 has no rational roots. Factors of degree 2 or more aren't searched for at degree 5, so it may factor further

   $\text{3x^5 − 17x + 7}$

4. Approximate roots of 3x^5 − 17x + 7, found numerically (Durand–Kerner iteration)

   $\text{x ≈ −1.6321, x ≈ −0.0988 ± 1.5587i, x ≈ 0.4139, x ≈ 1.4159}$

5. Put together the factors found. The last factor may break down further

   $\text{3x^5 − 17x + 7 = 3x^5 − 17x + 7}$

6. Factored form: −(x^2 + 17x − 12)

1. Write the polynomial in standard form, highest power first

   $\text{p(x) = −x^2 − 17x + 12}$

2. Take out the constant factor −1, leaving x^2 + 17x − 12

   $\text{−1(x^2 + 17x − 12)}$

3. By the rational root theorem, any rational root is ±(a factor of 12) ÷ (a factor of 1)

   $\text{1, −1, 2, −2, 3, −3, 4, −4, 6, −6, 12, −12}$

4. x^2 + 17x − 12 has no rational roots (its discriminant 337 is not a perfect square), so it can't be factored further over the rationals

   $\text{b² − 4ac = 337}$

5. Roots of x^2 + 17x − 12 by the quadratic formula (discriminant 337, not a perfect square, so the roots are irrational)

   $\text{x = (−17 ± √337)/2}$

6. Put the factors together. Multiplying them back out gives the original polynomial

   $\text{−x^2 − 17x + 12 = −(x^2 + 17x − 12)}$

7. Factored form: x^5 − x^2 − 10x + 3

1. Write the polynomial in standard form, highest power first

   $\text{p(x) = x^5 − x^2 − 10x + 3}$

2. By the rational root theorem, any rational root is ±(a factor of 3) ÷ (a factor of 1)

   $\text{1, −1, 3, −3}$

3. x^5 − x^2 − 10x + 3 has no rational roots. Factors of degree 2 or more aren't searched for at degree 5, so it may factor further

   $\text{x^5 − x^2 − 10x + 3}$

4. Approximate roots of x^5 − x^2 − 10x + 3, found numerically (Durand–Kerner iteration)

   $\text{x ≈ −1.7745, x ≈ −0.1499 ± 1.7968i, x ≈ 0.2917, x ≈ 1.7827}$

5. Put together the factors found. The last factor may break down further

   $\text{x^5 − x^2 − 10x + 3 = x^5 − x^2 − 10x + 3}$

8. Factored form: x^5 − 10x + 1

1. Write the polynomial in standard form, highest power first

   $\text{p(x) = x^5 − 10x + 1}$

2. By the rational root theorem, any rational root is ±(a factor of 1) ÷ (a factor of 1)

   $\text{1, −1}$

3. x^5 − 10x + 1 has no rational roots. Factors of degree 2 or more aren't searched for at degree 5, so it may factor further

   $\text{x^5 − 10x + 1}$

4. Approximate roots of x^5 − 10x + 1, found numerically (Durand–Kerner iteration)

   $\text{x ≈ −1.8024, x ≈ −0.025 ± 1.7792i, x ≈ 0.1, x ≈ 1.7523}$

5. Put together the factors found. The last factor may break down further

   $\text{x^5 − 10x + 1 = x^5 − 10x + 1}$

9. Factored form: 3x^6 − 2x^4 − 16x + 5

1. Write the polynomial in standard form, highest power first

   $\text{p(x) = 3x^6 − 2x^4 − 16x + 5}$

2. By the rational root theorem, any rational root is ±(a factor of 5) ÷ (a factor of 3)

   $\text{1/3, −1/3, 1, −1, 5/3, −5/3, 5, −5}$

3. 3x^6 − 2x^4 − 16x + 5 has no rational roots. Factors of degree 2 or more aren't searched for at degree 6, so it may factor further

   $\text{3x^6 − 2x^4 − 16x + 5}$

4. Approximate roots of 3x^6 − 2x^4 − 16x + 5, found numerically (Durand–Kerner iteration)

   $\text{x ≈ −1.2597 ± 0.7695i, x ≈ 0.3115, x ≈ 0.3847 ± 1.2486i, x ≈ 1.4385}$

5. Put together the factors found. The last factor may break down further

   $\text{3x^6 − 2x^4 − 16x + 5 = 3x^6 − 2x^4 − 16x + 5}$

10. Factored form: 2(x^5 − 3x^4 − 8x + 6)

1. Write the polynomial in standard form, highest power first

   $\text{p(x) = 2x^5 − 6x^4 − 16x + 12}$

2. Take out the constant factor 2, leaving x^5 − 3x^4 − 8x + 6

   $\text{2(x^5 − 3x^4 − 8x + 6)}$

3. By the rational root theorem, any rational root is ±(a factor of 6) ÷ (a factor of 1)

   $\text{1, −1, 2, −2, 3, −3, 6, −6}$

4. x^5 − 3x^4 − 8x + 6 has no rational roots. Factors of degree 2 or more aren't searched for at degree 5, so it may factor further

   $\text{x^5 − 3x^4 − 8x + 6}$

5. Approximate roots of x^5 − 3x^4 − 8x + 6, found numerically (Durand–Kerner iteration)

   $\text{x ≈ −1.4067, x ≈ 0.266 ± 1.3709i, x ≈ 0.686, x ≈ 3.1887}$

6. Put together the factors found. The last factor may break down further

   $\text{2x^5 − 6x^4 − 16x + 12 = 2(x^5 − 3x^4 − 8x + 6)}$