Question

If the polynomial 4x3- 16x2 +ax +7 is exactly divisible by x-1, then find the value of a, hence factorise the polynomial.

Answer 1

Its given that (x-1) is factor of 4x3-16x2+ax+7 So we vl put x=1 in equation V vl get 4-16+a+7=0 a=5

Answer 2

Answer:

Polynomial : $4x^3- 16x^2 +ax +7=0$

We are given that (x-1) is dividing exactly the given polynomial.

So, the roots of given equation :

x-1=0

x=1

Since x =1 is the root of given polynomial

So, it must satisfy the equation of given polynomial.

$4(1)^3- 16(1)^2 +a(1) +7=0$

$4-16+a(1) +7=0$

$-12+a(1) +7=0$

$-5+a=0$

$a=5$

So, the polynomial is $4x^3- 16x^2 +5x +7=0$

Divide $4x^3- 16x^2 +5x +7$ by (x-1)

$Dividend =( Divisor \times quotient)+Remainder$

$4x^3- 16x^2 +5x +7 =(x-1\times 4x^2-12x-7)+0$

quotient = $4x^2-12x-7$

factorize quotient

$4x^2-12x-7$

$4x^2-14x+2x-7$

$2x(2x+1)-7(2x+1)$

$(2x+1)(2x-7)$

So, the factors of polynomial are (x-1) , (2x+1) and (2x-7)