1. The problem is to factor and simplify the given polynomial expressions from problems 2 to 10. 2. We use common factoring techniques such as: - Difference of squares: $$a^2 - b^2 = (a-b)(a+b)$$ - Perfect square trinomials: $$a^2 \pm 2ab + b^2 = (a \pm b)^2$$ - Factoring by grouping - Sum and difference of cubes 3. Let's solve each problem step-by-step with intermediate work and explanations. **Problem 2:** ж) $$4a^2 - 20ab + 25b^2 - 36$$ Recognize $$4a^2 - 20ab + 25b^2 = (2a - 5b)^2$$, so expression is: $$ (2a - 5b)^2 - 6^2 $$ Use difference of squares: $$ (2a - 5b - 6)(2a - 5b + 6) $$ з) $$16m^2 - 8mn + n^2 - 49$$ Recognize $$16m^2 - 8mn + n^2 = (4m - n)^2$$, so: $$ (4m - n)^2 - 7^2 $$ Difference of squares: $$ (4m - n - 7)(4m - n + 7) $$ и) $$25x^2 - 4a^2 + 12ab - 9b^2$$ Group terms: $$ 25x^2 - (4a^2 - 12ab + 9b^2) $$ Recognize inside parentheses as perfect square: $$ 25x^2 - (2a - 3b)^2 $$ Difference of squares: $$ (5x - (2a - 3b))(5x + (2a - 3b)) = (5x - 2a + 3b)(5x + 2a - 3b) $$ к) $$9x^2 - 4y^2 + 4yz - z^2$$ Rewrite: $$ 9x^2 - (4y^2 - 4yz + z^2) $$ Inside parentheses is perfect square: $$ (2y - z)^2 $$ So: $$ 9x^2 - (2y - z)^2 $$ Difference of squares: $$ (3x - (2y - z))(3x + (2y - z)) = (3x - 2y + z)(3x + 2y - z) $$ **Problem 6:** а) $$a^2 - b^2 - a + b$$ Group: $$ (a^2 - b^2) - (a - b) $$ Factor difference of squares: $$ (a - b)(a + b) - (a - b) $$ Factor out $$ (a - b) $$: $$ (a - b)(a + b - 1) $$ б) $$x^2 - y^2 + x + y$$ Group: $$ (x^2 - y^2) + (x + y) $$ Factor difference of squares: $$ (x - y)(x + y) + (x + y) $$ Factor out $$ (x + y) $$: $$ (x + y)(x - y + 1) $$ в) $$m^3 - m^2 n - mn^2 + n^3$$ Group: $$ (m^3 - m^2 n) - (mn^2 - n^3) $$ Factor each group: $$ m^2(m - n) - n^2(m - n) $$ Factor out $$ (m - n) $$: $$ (m - n)(m^2 - n^2) $$ Factor difference of squares: $$ (m - n)(m - n)(m + n) = (m - n)^2 (m + n) $$ г) $$x^3 + x^2 y - xy^2 - y^3$$ Group: $$ (x^3 + x^2 y) - (xy^2 + y^3) $$ Factor each group: $$ x^2(x + y) - y^2(x + y) $$ Factor out $$ (x + y) $$: $$ (x + y)(x^2 - y^2) $$ Factor difference of squares: $$ (x + y)(x - y)(x + y) = (x + y)^2 (x - y) $$ д) $$a^2 + 2ab + b^2 - ac - bc$$ Group: $$ (a^2 + 2ab + b^2) - (ac + bc) $$ Recognize perfect square: $$ (a + b)^2 - c(a + b) $$ Factor out $$ (a + b) $$: $$ (a + b)(a + b - c) $$ е) $$xz - yz - x^2 + 2xy - y^2$$ Group: $$ (xz - yz) - (x^2 - 2xy + y^2) $$ Factor first group: $$ z(x - y) - (x - y)^2 $$ Rewrite: $$ z(x - y) - (x - y)^2 = (x - y)(z - (x - y)) = (x - y)(z - x + y) $$ ж) $$m^2 + 2mn + n^2 - p^2 + 2pq - q^2$$ Group: $$ (m^2 + 2mn + n^2) - (p^2 - 2pq + q^2) $$ Recognize perfect squares: $$ (m + n)^2 - (p - q)^2 $$ Difference of squares: $$ ((m + n) - (p - q))((m + n) + (p - q)) = (m + n - p + q)(m + n + p - q) $$ з) $$a^2 + 2ab + b^2 - c^2 - 2cd - d^2$$ Group: $$ (a^2 + 2ab + b^2) - (c^2 + 2cd + d^2) $$ Recognize perfect squares: $$ (a + b)^2 - (c + d)^2 $$ Difference of squares: $$ ((a + b) - (c + d))((a + b) + (c + d)) = (a + b - c - d)(a + b + c + d) $$ **Problem 7:** а) $$x^5 - x^3 + x^2 - 1$$ Group: $$ (x^5 - x^3) + (x^2 - 1) $$ Factor: $$ x^3(x^2 - 1) + (x^2 - 1) $$ Factor out $$ (x^2 - 1) $$: $$ (x^2 - 1)(x^3 + 1) $$ Factor difference of squares: $$ (x - 1)(x + 1)(x^3 + 1) $$ Factor sum of cubes: $$ (x - 1)(x + 1)(x + 1)(x^2 - x + 1) = (x - 1)(x + 1)^2 (x^2 - x + 1) $$ б) $$m^5 + m^3 - m^2 - 1$$ Group: $$ (m^5 + m^3) - (m^2 + 1) $$ Factor: $$ m^3(m^2 + 1) - (m^2 + 1) $$ Factor out $$ (m^2 + 1) $$: $$ (m^2 + 1)(m^3 - 1) $$ Factor difference of cubes: $$ (m^2 + 1)(m - 1)(m^2 + m + 1) $$ в) $$a^3 - 8 + 6a^2 - 12a$$ Rewrite: $$ a^3 + 6a^2 - 12a - 8 $$ Group: $$ (a^3 + 6a^2) - (12a + 8) $$ Factor: $$ a^2(a + 6) - 4(3a + 2) $$ No common factor, try rearranging: $$ (a^3 - 12a) + (6a^2 - 8) = a(a^2 - 12) + 2(3a^2 - 4) $$ No simple factorization, try grouping differently or use cubic formulas. г) $$p^3 + 8 + 6p^2 + 12p$$ Rewrite: $$ p^3 + 6p^2 + 12p + 8 $$ Group: $$ (p^3 + 6p^2) + (12p + 8) $$ Factor: $$ p^2(p + 6) + 4(3p + 2) $$ No common factor, try rearranging or sum of cubes: $$ (p + 2)^3 = p^3 + 6p^2 + 12p + 8 $$ So: $$ (p + 2)^3 $$ д) $$a^4 + a^3 + a + 1$$ Group: $$ (a^4 + a^3) + (a + 1) $$ Factor: $$ a^3(a + 1) + 1(a + 1) $$ Factor out $$ (a + 1) $$: $$ (a + 1)(a^3 + 1) $$ Factor sum of cubes: $$ (a + 1)(a + 1)(a^2 - a + 1) = (a + 1)^2 (a^2 - a + 1) $$ е) $$x^4 + x^3 - x - 1$$ Group: $$ (x^4 + x^3) - (x + 1) $$ Factor: $$ x^3(x + 1) - 1(x + 1) $$ Factor out $$ (x + 1) $$: $$ (x + 1)(x^3 - 1) $$ Factor difference of cubes: $$ (x + 1)(x - 1)(x^2 + x + 1) $$ ж) $$a^3 + a^2 b - ab^2 - b^3$$ Group: $$ (a^3 + a^2 b) - (ab^2 + b^3) $$ Factor: $$ a^2(a + b) - b^2(a + b) $$ Factor out $$ (a + b) $$: $$ (a + b)(a^2 - b^2) $$ Factor difference of squares: $$ (a + b)(a - b)(a + b) = (a + b)^2 (a - b) $$ з) $$x^3 - x^2 y - xy^2 + y^3$$ Group: $$ (x^3 - x^2 y) - (xy^2 - y^3) $$ Factor: $$ x^2(x - y) - y^2(x - y) $$ Factor out $$ (x - y) $$: $$ (x - y)(x^2 - y^2) $$ Factor difference of squares: $$ (x - y)(x - y)(x + y) = (x - y)^2 (x + y) $$ **Problem 8:** а) $$m^4 - n^4$$ Difference of squares: $$ (m^2 - n^2)(m^2 + n^2) $$ Further factor: $$ (m - n)(m + n)(m^2 + n^2) $$ б) $$a^6 - b^6$$ Difference of squares: $$ (a^3 - b^3)(a^3 + b^3) $$ Factor sum and difference of cubes: $$ (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2) $$ в) $$x^4 + x^3 + x + 1$$ Group: $$ (x^4 + x^3) + (x + 1) $$ Factor: $$ x^3(x + 1) + 1(x + 1) $$ Factor out $$ (x + 1) $$: $$ (x + 1)(x^3 + 1) $$ Factor sum of cubes: $$ (x + 1)(x + 1)(x^2 - x + 1) = (x + 1)^2 (x^2 - x + 1) $$ г) $$a^6 - a^4 + 2a^3 + 2a^2$$ Group: $$ a^4(a^2 - 1) + 2a^2(a + 1) $$ Rewrite: $$ a^4(a - 1)(a + 1) + 2a^2(a + 1) $$ Factor out $$ a^2(a + 1) $$: $$ a^2(a + 1)(a^2(a - 1) + 2) $$ д) $$ (a + b)^3 - (a - b)^3 $$ Use formula for difference of cubes: $$ (a + b - (a - b))((a + b)^2 + (a + b)(a - b) + (a - b)^2) $$ Simplify first factor: $$ 2b $$ Simplify second factor: $$ (a + b)^2 + (a + b)(a - b) + (a - b)^2 = 3a^2 + b^2 $$ So: $$ 2b(3a^2 + b^2) $$ е) $$ (a + b)^4 - (a - b)^4 $$ Difference of squares: $$ ((a + b)^2 - (a - b)^2)((a + b)^2 + (a - b)^2) $$ Simplify first factor: $$ 4ab \times 2 = 4ab \times 2 = 8ab $$ Simplify second factor: $$ 2(a^2 + b^2) $$ So: $$ 8ab(a^2 + b^2) $$ **Problem 9:** а) $$x^2 - 5x + 6$$ Factor: $$ (x - 2)(x - 3) $$ б) $$x^2 + 6x + 8$$ Factor: $$ (x + 2)(x + 4) $$ в) $$a^2 - 7ab + 12b^2$$ Factor: $$ (a - 3b)(a - 4b) $$ г) $$a^2 - 7ab + 10b^2$$ Factor: $$ (a - 5b)(a - 2b) $$ д) $$x^2 - x - 12$$ Factor: $$ (x - 4)(x + 3) $$ е) $$x^2 + x - 12$$ Factor: $$ (x + 4)(x - 3) $$ ж) $$a^2 - 3ab - 10b^2$$ Factor: $$ (a - 5b)(a + 2b) $$ з) $$a^2 + 2ab - 15b^2$$ Factor: $$ (a + 5b)(a - 3b) $$ и) $$2a^2 + 10a + 12$$ Factor out 2: $$ 2(a^2 + 5a + 6) $$ Factor inside: $$ 2(a + 2)(a + 3) $$ к) $$2x^2 + 14x + 24$$ Factor out 2: $$ 2(x^2 + 7x + 12) $$ Factor inside: $$ 2(x + 3)(x + 4) $$ л) $$2m^2 - 6m + 4$$ Factor out 2: $$ 2(m^2 - 3m + 2) $$ Factor inside: $$ 2(m - 1)(m - 2) $$ м) $$3p^2 + 27p + 54$$ Factor out 3: $$ 3(p^2 + 9p + 18) $$ Factor inside: $$ 3(p + 3)(p + 6) $$ **Problem 10:** а) $$a^8 + a^4 + 1$$ No simple factorization over integers. б) $$a^4 + a^2 b^2 + b^4$$ Recognize as sum of squares pattern, no simple factorization over reals. в) $$a^3 - 3a + 2$$ Try rational roots: test $$a=1$$: $$1 - 3 + 2 = 0$$ So factor: $$ (a - 1)(a^2 + a - 2) $$ Factor quadratic: $$ (a - 1)(a - 1)(a + 2) = (a - 1)^2 (a + 2) $$ Final answers are factored forms as above.