Question

1. factor completely each of the following expressions.

(a) \\(12 + x - x^2\\)
(b) \\(2p^3 - 12p^2 + 16p\\)
(c) \\(8x^2 - 10x - 3\\)
(d) \\(63x^2 - 33x - 6\\)
(e) \\(49y^2 - 81\\)
(f) \\(25p^2 - 30p + 9\\)
(g) \\(10a^2 + 11ab - 6b^2\\)
(h) \\(200x^2 - 18\\)
(i) \\(16u^2 + 20u - 6\\)
(j) \\(w^2 - 3w - 28\\)
(k) \\(-18x^3 - 15x^2 + 18x\\)
(l) \\(3y^2 - 9y - 30\\)

Explanation:

Factor expressions (a) to (c)

Using the Polynomial Factoring knowledge point, we factor the first three quadratic expressions.
For (a):

$$ 12 + x - x^2 = -(x^2 - x - 12) = -(x - 4)(x + 3) = (4 - x)(3 + x) $$

For (b), we first extract the greatest common factor \(2p\):

$$ 2p^3 - 12p^2 + 16p = 2p(p^2 - 6p + 8) = 2p(p - 4)(p - 2) $$

For (c), we use splitting the middle term:

$$ 8x^2 - 10x - 3 = 8x^2 - 12x + 2x - 3 = 4x(2x - 3) + 1(2x - 3) = (4x + 1)(2x - 3) $$

Factor expressions (d) to (f)

Using the Polynomial Factoring and Difference of Squares knowledge points:
For (d), we factor out the common factor \(3\):

$$ 63x^2 - 33x - 6 = 3(21x^2 - 11x - 2) = 3(21x^2 - 14x + 3x - 2) = 3[7x(3x - 2) + 1(3x - 2)] = 3(7x + 1)(3x - 2) $$

For (e), we apply the difference of squares:

$$ 49y^2 - 81 = (7y)^2 - 9^2 = (7y - 9)(7y + 9) $$

For (f), we recognize a perfect square trinomial:

$$ 25p^2 - 30p + 9 = (5p)^2 - 2(5p)(3) + 3^2 = (5p - 3)^2 $$

Factor expressions (g) to (i)

Using the Polynomial Factoring and Difference of Squares knowledge points:
For (g), we factor the two-variable quadratic:

$$ 10a^2 + 11ab - 6b^2 = 10a^2 + 15ab - 4ab - 6b^2 = 5a(2a + 3b) - 2b(2a + 3b) = (5a - 2b)(2a + 3b) $$

For (h), we factor out the common factor \(2\):

$$ 200x^2 - 18 = 2(100x^2 - 9) = 2[(10x)^2 - 3^2] = 2(10x - 3)(10x + 3) $$

For (i), we factor out the common factor \(2\):

$$ 16u^2 + 20u - 6 = 2(8u^2 + 10u - 3) = 2(8u^2 + 12u - 2u - 3) = 2[4u(2u + 3) - 1(2u + 3)] = 2(4u - 1)(2u + 3) $$

Factor expressions (j) to (l)

Using the Polynomial Factoring knowledge point:
For (j):

$$ w^2 - 3w - 28 = (w - 7)(w + 4) $$

For (k), we factor out the common factor \(-3x\):

$$ -18x^3 - 15x^2 + 18x = -3x(6x^2 + 5x - 6) = -3x(6x^2 + 9x - 4x - 6) = -3x[3x(2x + 3) - 2(2x + 3)] = -3x(3x - 2)(2x + 3) $$

For (l), we factor out the common factor \(3\):

$$ 3y^2 - 9y - 30 = 3(y^2 - 3y - 10) = 3(y - 5)(y + 2) $$

Answer:

No.	Problem	Answer
(b)	\(2p^3 - 12p^2 + 16p\)	\(2p(p - 4)(p - 2)\)
(c)	\(8x^2 - 10x - 3\)	\((4x + 1)(2x - 3)\)
(d)	\(63x^2 - 33x - 6\)	\(3(7x + 1)(3x - 2)\)
(e)	\(49y^2 - 81\)	\((7y - 9)(7y + 9)\)
(f)	\(25p^2 - 30p + 9\)	\((5p - 3)^2\)
(g)	\(10a^2 + 11ab - 6b^2\)	\((5a - 2b)(2a + 3b)\)
(h)	\(200x^2 - 18\)	\(2(10x - 3)(10x + 3)\)
(i)	\(16u^2 + 20u - 6\)	\(2(4u - 1)(2u + 3)\)
(j)	\(w^2 - 3w - 28\)	\((w - 7)(w + 4)\)
(k)	\(-18x^3 - 15x^2 + 18x\)	\(-3x(3x - 2)(2x + 3)\)
(l)	\(3y^2 - 9y - 30\)	\(3(y - 5)(y + 2)\)