Question

a) Without actual division, prove that 2x^4 – 6x^3 + 3x^2 + 3x - 2 is exactly divisible by
x² – 3x +2.
​

Answer 1

Required Knowledge

• Polynomial division

Consider dividing a polynomial $F(x)$ by $P(x)$. The quotient is $Q(x)$ and the remainder is $R(x)$.

$\implies F(x)=P(x)\cdot Q(x)+R(x)$

• Process of division

The process of division is elimination of the highest degree. So, when the divisor of degree $n$ has divided the polynomial, the remainder is at most $n-1$ degree.

Solution

Consider,

$2x^4-6x^3+3x^2+3x-2=(x-1)(x-2)Q(x)+R(x)$

Since the divisor has a degree of 2, the remainder has a degree of at most 1.

So assume,

$R(x)$ is a linear polynomial. Let $R(x)=ax+b$.

Then substituting $x=1$ or $x=2$ gives,

$x=1\implies 2-6+3+3-2=R(1)$

$x=2\implies 32-48+12+6-2=R(2)$

So this gives

$R(1)=a+b=0$

$R(2)=2a+b=0$

That results to $\boxed{R(x)=0}$. And this states the polynomial is exactly divisible by $(x-1)(x-2)$.

Answer 2

Required Answer :-

According to the question

g(x) = x² - 3x + 2

g(x) = x² - (2x + x) + 2

g(x) = x² - 2x - x + 2

g(x) = x(x - 2) - 1(x - 2)

g(x) = (x - 2),(x - 1)

x = 2

or

x = 1

By putting them

f(x)  = 2(2)⁴ - 6(2)³ + 3(2)² + 3(2) - 2

f(2) = 2(16) - 6(8) +3(4) - 2

f(2) = 32 - 48 + 12 + 6 -  2

f(2) = 0

Now

f(1) = 2(1)⁴ - 6(1)³ + 3(1)² + 3(1) - 2

f(1) = 2 - 6 + 3 + 3 - 2

f(1) = -6 + 6

f(1) = 0

Hence,

Proved