using factor theorem select the factor of the polynomial x3-7x2+16x-12 ? (a) x+2 (b) x-3 (c) x-2 (d) x-4
Answers
Step-by-step explanation:
Given :-
The polynomial x³-7x²+16x-12
To find :-
Using factor theorem select the factor of the following (a) x+2 (b) x-3 (c) x-2 (d) x-4 ?
Solution :-
Given polynomial is x³-7x²+16x-12
Let P(x) = x³-7x²+16x-12
We know that
Factor Theorem : Let P(x) be a polynomial of the degree greater than or equal to 1 and (x-a) is another linear polynomial if (x-a) is a factor of P(x) then P(a) = 0 vice-versa.
a)Checking x+2 is a factor or not :-
If (x+2) is a factor of P (x) then P(-2) = 0
Put x = -2 then
=> P(-2) =(-2)³-7(-2)²+16(-2)-12
=> P(-2) = -8-7(4)+(-32)-12
=> P(-2) = -8-28-32-12
=> P(-2) = -80
=> P(-2)≠0
(x+2) is not a factor of P(x).
b) Checking x-3 is a factor or not :-
If (x-3) is a factor of P (x) then P(3) = 0
Put x = 3 then
=> P(3) =(3)³-7(3)²+16(3)-12
=> P(3) = 27-7(9)+(48)-12
=> P(3) = 27-63+48-12
=> P(3) = 75-75
=> P(3) = 0
(x-3) is a factor of P(x).
c) Checking x-2 is a factor or not :-
If (x-2) is a factor of P (x) then P(2) = 0
Put x = 2 then
=> P(2) =(2)³-7(2)²+16(2)-12
=> P(2) = 8-7(4)+(32)-12
=> P(2) = 8-28+32-12
=> P(2) = 40-40
=> P(2) = 0
(x-2) is a factor of P(x).
d) Checking x-4 is a factor or not :-
If (x-4) is a factor of P (x) then P(4) = 0
Put x = 4 then
=> P(4) =(4)³-7(4)²+16(4)-12
=> P(4) = 64-7(16)+(64)-12
=> P(4) = 64-118+64-12
=> P(4) = 128-130
=> P(4)= -2
=> P(4) ≠ 0
(x-4) is not a factor of P(x).
Answer:-
b)x-3 c)x-2 are the factors of the given Polynomial.
Used formulae :-
Factor Theorem : Let P(x) be a polynomial of the degree greater than or equal to 1 and (x-a) is another linear polynomial if (x-a) is a factor of P(x) then P(a) = 0 vice-versa.
Answer:
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