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X-box Factoring

X-box Factoring. X- Box. Trinomial (Quadratic Equation). Product of a & c. ax 2 + bx + c. Fill the 2 empty sides with 2 numbers that are factors of ‘a·c’ and add to give you ‘b’. b. X- Box. Trinomial (Quadratic Equation). x 2 + 9x + 20. 20. 5.

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X-box Factoring

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  1. X-box Factoring

  2. X- Box Trinomial (Quadratic Equation) Product of a & c ax2 + bx + c Fill the 2 empty sides with 2 numbers that are factors of ‘a·c’ and add to give you ‘b’. b

  3. X- Box Trinomial (Quadratic Equation) x2 + 9x + 20 20 5 Fill the 2 empty sides with 2 numbers that are factors of ‘a·c’ and add to give you ‘b’. 4 9

  4. X- Box Trinomial (Quadratic Equation) 2x2 -x - 21 -42 Fill the 2 empty sides with 2 numbers that are factors of ‘a·c’ and add to give you ‘b’. -7 6 -1

  5. X-box Factoring • This is a guaranteed method for factoring quadratic equations—no guessing necessary! • We will learn how to factor quadratic equations using the x-box method

  6. LET’S TRY IT! Students apply basic factoring techniques to second- and simple third-degree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials. Objective: I can use the x-box method to factor non-prime trinomials.

  7. Factor the x-box way Example: Factor x2 -3x -10 (1)(-10)= -10 x -5 x2 -5x x GCF -5 2 -3 -10 2x +2 GCF GCF GCF x2 -3x -10 = (x-5)(x+2)

  8. Factor the x-box way y = ax2 + bx + c Base 1 Base 2 Product ac=mn First and Last Coefficients 1st Term Factor n GCF n m Middle Last term Factor m b=m+n Sum Height

  9. Factor the x-box way Example: Factor 3x2 -13x -10 x -5 -30 3x 3x2 -15x 2 -15 -13 -10 2x +2 3x2 -13x -10 = (x-5)(3x+2)

  10. -12 4 Examples Factor using the x-box method. 1. x2 + 4x – 12 a) b) x +6 x26x -2x -12 x 6 -2 -2 Solution: x2 + 4x – 12 = (x + 6)(x - 2)

  11. Examples continued 2. x2 - 9x + 20 a) b) x -4 -4-5 20 -9 x2-4x -5x 20 x -5 Solution: x2 - 9x + 20 =(x - 4)(x - 5)

  12. Examples continued 3. 2x2 - 5x - 7 a) b) 2x -7 -72 -14 -5 x 2x2-7x 2x -7 +1 Solution: 2x2 - 5x – 7 = (2x - 7)(x + 1)

  13. Examples continued 3. 15x2 + 7x - 2 a) b) 3x +2 -30 7 5x 15x210x -3x -2 10-3 -1 Solution: 15x2 + 7x – 2 = (3x + 2)(5x - 1)

  14. Extra Practice • x2 +4x -32 • 4x2 +4x -3 • 3x2 + 11x – 20

  15. Reminder!! Don’t forget to check your answer by multiplying!

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