Question

iew
factoring:

1. $9x^2 - 1$
2. $x^2 - 16$
3. $-x^2 + 16$
4. $12x^2 - 75$
5. $-98 + 2x^2$
6. $2x^2 + 8x + 6$
7. $4x^2 + 16x + 16$
8. $14x^2 + 7x - 21$
9. $5x^2 + 2x + 7$

solving by factoring:

1. $(x + 1)(x - 5) = 0$
2. $(x + 1)(x + 2) = 0$
3. $(4x + 5)(x + 1) = 0$
4. $x^2 - 11x + 19 = -5$
5. $8x^2 + 18x + 9 = 0$
6. $x^2 + 7x + 15 = 5$
7. $x^2 - 10x - 22 = 2$
8. $x^2 + 3x - 12 = 6$
9. $6x^2 - 18x - 18 = 6$
10. $7x^2 - 14x = -7$
11. $x^2 + 8x = -15$
12. $x^2 + 5x - 35 = 3x$
13. $5x^2 - 44x + 120 = -30 + 11x$
14. $-4x^2 - 8x - 3 = -3 - 5x^2$
15. $3x^2 - 16x - 7 = 5$
16. $6x^2 - 13x + 3 = -3$
17. $2x^2 - 8 = 0$
18. $7x^2 + 2x = 0$

Explanation:

Step1: Factor \(9x^2 - 1\)

Notice that \(9x^2=(3x)^2\) and \(1 = 1^2\). This is a difference of squares, which factors as \(a^2 - b^2=(a - b)(a + b)\). So \(9x^2 - 1=(3x - 1)(3x + 1)\).

Step2: Factor \(x^2 - 16\)

\(x^2 - 16=x^2 - 4^2\), using the difference of squares formula, it factors to \((x - 4)(x + 4)\).

Step3: Factor \(-x^2 + 16\)

Rewrite as \(16 - x^2=4^2 - x^2\), then factor as \((4 - x)(4 + x)\) (or \(-(x - 4)(x + 4)\) if we factor out the negative first, but \((4 - x)(4 + x)\) is also correct).

Step4: Factor \(12x^2 - 75\)

First, factor out the greatest common factor (GCF) of 3: \(12x^2 - 75 = 3(4x^2 - 25)\). Then \(4x^2 - 25=(2x)^2 - 5^2\), so it factors to \(3(2x - 5)(2x + 5)\).

Step5: Factor \(-98 + 2x^2\)

Factor out the GCF of 2: \(2(x^2 - 49)\). Then \(x^2 - 49=x^2 - 7^2\), so it factors to \(2(x - 7)(x + 7)\).

Step6: Factor \(2x^2 + 8x + 6\)

Factor out the GCF of 2: \(2(x^2 + 4x + 3)\). Now factor the quadratic inside the parentheses. We need two numbers that multiply to 3 and add to 4, which are 1 and 3. So \(x^2 + 4x + 3=(x + 1)(x + 3)\), and the full factorization is \(2(x + 1)(x + 3)\).

Step7: Factor \(4x^2 + 16x + 16\)

Factor out the GCF of 4: \(4(x^2 + 4x + 4)\). The quadratic \(x^2 + 4x + 4\) is a perfect square, \((x + 2)^2\), so the full factorization is \(4(x + 2)^2\).

Step8: Factor \(14x^2 + 7x - 21\)

Factor out the GCF of 7: \(7(2x^2 + x - 3)\). Now factor the quadratic \(2x^2 + x - 3\). We need two numbers that multiply to \(2\times(-3)=-6\) and add to 1, which are 3 and -2. Rewrite the middle term: \(2x^2 + 3x - 2x - 3\). Group and factor: \(x(2x + 3)-1(2x + 3)=(2x + 3)(x - 1)\). So the full factorization is \(7(2x + 3)(x - 1)\).

Step9: Check \(5x^2 + 2x + 7\)

For a quadratic \(ax^2 + bx + c\), the discriminant is \(b^2 - 4ac=(2)^2 - 4\times5\times7 = 4 - 140=-136\), which is negative, so it does not factor over the real numbers.

Answer:

1. \(\boldsymbol{(3x - 1)(3x + 1)}\)
2. \(\boldsymbol{(x - 4)(x + 4)}\)
3. \(\boldsymbol{(4 - x)(4 + x)}\) (or \(\boldsymbol{-(x - 4)(x + 4)}\))
4. \(\boldsymbol{3(2x - 5)(2x + 5)}\)
5. \(\boldsymbol{2(x - 7)(x + 7)}\)
6. \(\boldsymbol{2(x + 1)(x + 3)}\)
7. \(\boldsymbol{4(x + 2)^2}\)
8. \(\boldsymbol{7(2x + 3)(x - 1)}\)
9. Does not factor over \(\mathbb{R}\)