Question

The root of the equation

y/3 – 7 = 11

n 2x + 3 = 2(x – 4)

Answer 1

Step-by-step explanation:

p(x)=x3−6x2+15x−8 ∴ degree of p(x) is 3.

g(x)=x−2 ∴ degree of g(x) is 1.

∴ degree of quotient q(x)=3−1=2 and degree of remainder r(x) is zero.

Let, q(x)=ax2+bx+c (Polynomial of degree 2) and r(x)=k (constant polynomial)

By using division algorithm, we have

p(x)=[g(x)×q(x)+r(x)]=x3−6x2+15x−8=(x−2)(ax2+bx+c)+k

=ax3+bx2+cx−2ax2−2bx−2c+k

∴x3−6x2+15x−8=ax3+(b−2a)x2+(c−2b)x−2c+k

We have cubic polynomials on both the sides of the equation.

∴ Let us compare the coefficients of x3,x2,x and k to get the values of a,b,c.

1=a coefficient of x