Question

check whether polynomial x-1,is a factor of the polynomial x³-8x²+19x-12.verify by division algorithm​

Answer 1

Step-by-step explanation:

Use remainder factor theorem

x-1=0

x=1

put the value

x3-8x2+19x-12=0

LHS=

1-8+19-12

=0=RHS

Hence x-1 is a factor of the given expression

Answer 2

(x-1) is a factor of polynomial $p(x) = x^3-8x^2+19x-12$

Step-by-step explanation:

We are given the following polynomial in the question:

$p(x) = x^3-8x^2+19x-12$

We have to check whether (x-1) is a factor of given polynomial or nor.

Put (x-1) = 0

$x-1 =0\\x = 1$

Now, we check if x = 1 is a root of given polynomial or not.

Putting x = 1, we get,

$p(1) = (1)^3-8(1)^2+19(1)-12\\p(1) = 0$

Thus, x = 1 is a root of the given polynomial.

Hence, (x-1) is a factor of polynomial $p(x) = x^3-8x^2+19x-12$

Division algorithm:

$p(x) = x^3-8x^2+19x-12 = (x^2-7x+12)(x-1)\\p(x) = q(x) (x-1) + 0\\q(x) = x^2-7x + 12$

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