Question

write the second degree polynomial x^2- 16 as a product of two first degree polynomials​

Answer 1

Step-by-step explanation:

(a) x

(a) x 2

(a) x 2 −3x+2=x

(a) x 2 −3x+2=x 2

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)(b) Since x

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)(b) Since x 2

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)(b) Since x 2 −3x+k can be written as the product of two first degree polynomials,

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)(b) Since x 2 −3x+k can be written as the product of two first degree polynomials,b

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)(b) Since x 2 −3x+k can be written as the product of two first degree polynomials,b 2

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)(b) Since x 2 −3x+k can be written as the product of two first degree polynomials,b 2 −4ac=0

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)(b) Since x 2 −3x+k can be written as the product of two first degree polynomials,b 2 −4ac=0(3)

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)(b) Since x 2 −3x+k can be written as the product of two first degree polynomials,b 2 −4ac=0(3) 2

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)(b) Since x 2 −3x+k can be written as the product of two first degree polynomials,b 2 −4ac=0(3) 2 −4(1)(k)=0⇒9−4k=0

(a) x 2 −3x+2=x 2 −2x−x+2=x(x−2)+1(x−2)=(x+1)(x−2)(b) Since x 2 −3x+k can be written as the product of two first degree polynomials,b 2 −4ac=0(3) 2 −4(1)(k)=0⇒9−4k=0k=9/4

Answer 2

Answer:

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