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Answer:
- zeros are {-5, -2, 2/3}
- zeros are {-5/3, 1/4, 5}
Step-by-step explanation:
"17." (the one before 18) Putting x=1 into the equation gives ...
... p(1) = 38+c
We're told this is 18, so we have ...
... 38+c = 18
... c = -20
Now, we know we have polynomial p(x) = 3x³+19x²+16x-20, and we know one root is x=-5. Using synthetic division to divide out the factor (x+5), we have the remaining quadratic factor (3x² +4x -4).
This can be factored by looking for factors of 3×(-4) that sum to 4. Those would be 6 and -2. If we call these "r" and "s", we can write the factored form of the quadratic as ...
... (3x+r)(3x+s)/3 = (3x+6)(3x-2)/3 = (x+2)(3x-2)
We have just factored the quadratic term. The entire p(x) is then factored as ...
... p(x) = 3x³ +19x² +16x -20 = (3x-2)(x+2)(x+5)
The roots are the values of x that make these factors zero:
... {2/3, -2, -5}
18. Using synthetic division to divide the polynomial by (x-5), we find the remainder is 5c+450. In order for that to be zero, the value of c must be -90. Then we have ...
... p(x) = 12x³ -43x² -90x +25
Checking the value of p(-2), we find it is -63, as advertised.
The same synthetic division that told us c=-90 gives the quadratic factor as (12x²+17x-5). This can be factored by looking for factors of 12×(-5) that sum to 17. 20 and -3 are those factors. So, we can write the quadratic as ...
... (12x +20)(12x -3)/12 = (3x +5)(4x -1)
Then the factorization of p(x) is ...
... p(x) = 12x³ -43x² -90x +25 = (3x +5)(4x -1)(x -5)
The roots are the values of x that make these factors zero:
... {-5/3, 1/4, 5}
