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1. $8x^2 + 12x$ 2. $15x^2 - 60$ 3. $x^2 - 81$ 4. $9x^2 - 25y^2$ 5. $x^2…

Question

  1. $8x^2 + 12x$
  2. $15x^2 - 60$
  3. $x^2 - 81$
  4. $9x^2 - 25y^2$
  5. $x^2 + 12x + 36$
  6. $4x^2 - 20x + 25$
  7. $2x^2 + 14x + 24$
  8. $x^3 + 2x^2 + 3x + 6$
  9. $6x^2 + 7x - 3$
  10. $12x^2 - 3x - 15$

Explanation:

Step1: Factor \(8x^2 + 12x\)

Find the GCF of \(8x^2\) and \(12x\). GCF of 8 and 12 is 4, GCF of \(x^2\) and \(x\) is \(x\). So GCF is \(4x\).
\(8x^2 + 12x = 4x(2x + 3)\)

Step2: Factor \(15x^2 - 60\)

GCF of 15 and 60 is 15.
\(15x^2 - 60 = 15(x^2 - 4)\). Then factor \(x^2 - 4\) as difference of squares: \(x^2 - 4=(x + 2)(x - 2)\). So \(15x^2 - 60 = 15(x + 2)(x - 2)\)

Step3: Factor \(x^2 - 81\)

This is difference of squares (\(a^2 - b^2=(a + b)(a - b)\)), where \(a = x\), \(b = 9\).
\(x^2 - 81=(x + 9)(x - 9)\)

Step4: Factor \(9x^2 - 25y^2\)

Difference of squares, \(a = 3x\), \(b = 5y\).
\(9x^2 - 25y^2=(3x + 5y)(3x - 5y)\)

Step5: Factor \(x^2 + 12x + 36\)

Perfect square trinomial: \(a^2 + 2ab + b^2=(a + b)^2\), here \(a = x\), \(2ab = 12x\) (so \(b = 6\)), \(b^2 = 36\).
\(x^2 + 12x + 36=(x + 6)^2\)

Step6: Factor \(4x^2 - 20x + 25\)

Perfect square trinomial: \(a^2 - 2ab + b^2=(a - b)^2\), \(a = 2x\), \(2ab = 20x\) (so \(b = 5\)), \(b^2 = 25\).
\(4x^2 - 20x + 25=(2x - 5)^2\)

Step7: Factor \(2x^2 + 14x + 24\)

First GCF: GCF of 2,14,24 is 2. \(2(x^2 + 7x + 12)\). Then factor \(x^2 + 7x + 12\): find two numbers that multiply to 12 and add to 7 (3 and 4). So \(x^2 + 7x + 12=(x + 3)(x + 4)\). Thus \(2x^2 + 14x + 24 = 2(x + 3)(x + 4)\)

Step8: Factor \(x^3 + 2x^2 + 3x + 6\)

Grouping: group first two and last two terms. \((x^3 + 2x^2)+(3x + 6)=x^2(x + 2)+3(x + 2)=(x^2 + 3)(x + 2)\)

Step9: Factor \(6x^2 + 7x - 3\)

Use AC method: \(a = 6\), \(b = 7\), \(c=-3\). \(ac=-18\). Find two numbers that multiply to -18 and add to 7 (9 and -2). Rewrite middle term: \(6x^2 + 9x - 2x - 3\). Group: \((6x^2 + 9x)-(2x + 3)=3x(2x + 3)-1(2x + 3)=(3x - 1)(2x + 3)\)

Step10: Factor \(12x^2 - 3x - 15\)

GCF: GCF of 12,-3,-15 is 3. \(3(4x^2 - x - 5)\). Factor \(4x^2 - x - 5\): AC method, \(ac=-20\), numbers 4 and -5 (4-5=-1). Rewrite: \(4x^2 + 4x - 5x - 5=4x(x + 1)-5(x + 1)=(4x - 5)(x + 1)\). So \(12x^2 - 3x - 15 = 3(4x - 5)(x + 1)\)

Answer:

s (for each problem):

  1. \(4x(2x + 3)\)
  2. \(15(x + 2)(x - 2)\)
  3. \((x + 9)(x - 9)\)
  4. \((3x + 5y)(3x - 5y)\)
  5. \((x + 6)^2\)
  6. \((2x - 5)^2\)
  7. \(2(x + 3)(x + 4)\)
  8. \((x^2 + 3)(x + 2)\)
  9. \((3x - 1)(2x + 3)\)
  10. \(3(4x - 5)(x + 1)\)