Mathematics, 12.11.2019 20:31, quay39
Davide wants to find the zeros of the polynomial p(x)=x3β2x2β9x+18. his work is shown below.
step 1: p(x)=(x+3)(xβ3)(xβ2)
step 2: x+3=0
xβ3=0xβ2=0
step 3: x=2, x=3, x=β3
explain why davideβs work shows that 2, 3, and β3 are zeros of p(x) by matching each step with the appropriate justification.
step 1:
step 2:
step 3:
(a) davide solves for x in each of the equations from step 2. the zeros of a polynomial are the x -values for which p(x)=0 .
(b) davide gets rid of the x in each of the equations from step 2 by setting x=0 . the zeros of a polynomial occur when x=0 .
(c) davide factors the polynomial p(x) into linear factors using factoring by grouping and the difference of squares.
(d) davide factors the polynomial p(x) into linear factors using the difference of cubes method.
(e) davide sets each linear factor equal to 0 because the x -value for the zero of a polynomial is 0 .
(f) davide sets each linear factor equal to 0 because of the zero product property.
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Davide wants to find the zeros of the polynomial p(x)=x3β2x2β9x+18. his work is shown below.
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