Math, asked by yashisharma14402, 4 hours ago

Check whether polynomial ( x – 1) is a factor of the polynomial (x3– 8x2+19x –12).Verify by division algorithm.

Answers

Answered by ks854092
0

Answer:

sorry I don't know if you have any questions or concerns please

Answered by tennetiraj86
7

Step-by-step explanation:

Given :-

The polynomial x³– 8x²+19x –12

To find :-

Check whether polynomial (x-1) is a factor of the polynomial (x³- 8x²+19x -12).

Verify by division algorithm.

Solution :-

Given Cubic Polynomial is

p(x) =x³-8x²+19x -12

Given linear polynomial = (x-1)

We know that

Factor Theorem : If (x-a) is a factor of p(x) then P(a) = 0

If (x-1) is a factor of p(x) then p(1) = 0

=> (1)³-8(1)²+19(1)-12

=> 1-8(1)+19-12

=> 1-8+19-12

=> (1+19)+(-8-12)

=> 20+(-20)

=> 20-20

=> 0

So we have p(1) = 0

Therefore, (x-1) is a factor of p(x).

Verification:-

Division Algorithm on Polynomials is

p(x) = g(x)×q(x)+r(x)

Where, p(x) is the given Polynomial or dividend

g(x) = divisor

q(x) = quotient

r(x) = remainder

(x-1) is a factor of p(x) then the remainder is 0

x-1)x³-8x²+19x-12(x²-7x+12

x³-x²

(-)

_____________

0 -7x²+19x

-7x²+7x

(+)

______________

0 +12x-12

12x-12

_______________

0

_______________

So , the remainder is zero .

and

p(x) = g(x)×q(x)+r(x)

=>(x-1) (x²-7x+12)+0

=> x(x²-7x+12)-1(x²-7x+12)

=> x³-7x²+12x-x²+7x-12

=> x³-8x²+19x-12

Verified the given relations in the given problem.

Used formulae:-

Division Algorithm on Polynomials is

p(x) = g(x)×q(x)+r(x)

Where, p(x) is the given Polynomial or dividend

g(x) = divisor

q(x) = quotient

r(x) = remainder

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.

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