$$ \begin{array}{ll} ext { (f) } & (x-1)^{3}-27 \ ext { (h) } 2 x^{3}+10 x^{2}+x+5 \ ext { (j) } & 4 x^{3}+12 x^{2}-x-3 \ ext { (l) } & 16 x^{3}+16\end{array} $$

Question

$$ \begin{array}{ll}	ext { (f) } & (x-1)^{3}-27 \ 	ext { (h) } 2 x^{3}+10 x^{2}+x+5 \ 	ext { (j) } & 4 x^{3}+12 x^{2}-x-3 \ 	ext { (l) } & 16 x^{3}+16\end{array} $$

UpStudy AI · Accepted Answer

**(f) Factorize $$ (x-1)^3 - 27 $$:**

1. Recognize that $$27 = 3^3$$. Thus, we have
   $$
   (x-1)^3 - 3^3.
   $$
2. Apply the difference of cubes formula:
   $$
   a^3 - b^3 = (a-b)(a^2 + ab + b^2),
   $$
   with $$ a = x-1 $$ and $$ b = 3 $$. Then,
   $$
   (x-1)^3 - 3^3 = \big((x-1)-3\big) \Big[(x-1)^2 + 3(x-1) + 9\Big].
   $$
3. Simplify the first factor:
   $$
   (x-1)-3 = x-4.
   $$
4. Expand and simplify the second factor:
   $$
   (x-1)^2 = x^2-2x+1,
   $$
   $$
   3(x-1) = 3x-3.
   $$
   Adding these with 9 gives:
   $$
   x^2-2x+1 + 3x-3 + 9 = x^2 + x + 7.
   $$
5. Final factorization:
   $$
   (x-1)^3 - 27 = (x-4)(x^2+x+7).
   $$

---

**(h) Factorize $$ 2x^3+10x^2+x+5 $$:**

1. Group the terms:
   $$
   (2x^3+10x^2) + (x+5).
   $$
2. Factor out common factors from each group:
   $$
   2x^3+10x^2 = 2x^2(x+5),
   $$
   $$
   x+5 = 1(x+5).
   $$
3. Factor out the common binomial $$ (x+5) $$:
   $$
   2x^2(x+5) + 1(x+5) = (x+5)(2x^2+1).
   $$

---

**(j) Factorize $$ 4x^3+12x^2-x-3 $$:**

1. Group the terms:
   $$
   (4x^3+12x^2) + (-x-3).
   $$
2. Factor from the first group:
   $$
   4x^3+12x^2 = 4x^2(x+3).
   $$
3. Factor the second group:
   $$
   -x-3 = -1(x+3).
   $$
4. Factor out the common binomial $$ (x+3) $$:
   $$
   4x^2(x+3) - 1(x+3) = (x+3)(4x^2-1).
   $$
5. Recognize that $$ 4x^2-1 $$ is a difference of squares:
   $$
   4x^2-1 = (2x)^2 - 1^2 = (2x-1)(2x+1).
   $$
6. Final factorization:
   $$
   4x^3+12x^2-x-3 = (x+3)(2x-1)(2x+1).
   $$

---

**(l) Factorize $$ 16x^3+16 $$:**

1. Factor out the greatest common factor (GCF) of 16:
   $$
   16x^3+16 = 16(x^3+1).
   $$
2. Recognize that $$ x^3+1 $$ is a sum of cubes, since $$ 1 = 1^3 $$. Use the sum of cubes formula:
   $$
   a^3 + b^3 = (a+b)(a^2-ab+b^2),
   $$
   with $$ a = x $$ and $$ b = 1 $$. Then,
   $$
   x^3+1 = (x+1)(x^2-x+1).
   $$
3. Final factorization:
   $$
   16x^3+16 = 16(x+1)(x^2-x+1).
   $$
**(f) $$ (x-1)^3 - 27 = (x-4)(x^2 + x + 7) $$** **(h) $$ 2x^3 + 10x^2 + x + 5 = (x + 5)(2x^2 + 1) $$** **(j) $$ 4x^3 + 12x^2 - x - 3 = (x + 3)(2x - 1)(2x + 1) $$** **(l) $$ 16x^3 + 16 = 16(x + 1)(x^2 - x + 1) $$**

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Algebra Questions