Question

If both (x-3) and (3x-1) are factors of ax^2+4x+b,prove that a-b=0

Answer 1

Step-by-step explanation:

If both (x - 3) and (3x - 1) are factors of ax² + 4x + b, then when they divide the polynomial, the remainder so obtained should be 0.

So,

We can use the Remainder Theorem to find the remainders

x - 3 = 0

x = 3

and

3x - 1 = 0

x = 1/3

Hence,

p(3) = p(1/3) = 0 (i)

Now,

p(3) = a(3²) + 4(3) + b

=9a + 12 + b = 0 (ii)

and,

p(1/3) = a(1/3)² + 4(1/3) + b

=a/9 + 4/3 + b = 0 (iii)

Since p(3) = p(1/3),

9a + 12 + b = a/9 + 4/3 + b

$9a + 12 = \frac{a +12}{9}$ [∵ b and b cancel out]

9(9a + 12) = a + 12

81a + 108 = a + 12

81a - a = 12 - 108

80a = -96

a = -96/80

= -6/5

b = -9a - 12 (From ii)

= 9(-6/5) - 12

= -54/5 + 12

= -6/5

--------------------------------------

a = -6/5 and b = -6/5

a - b = -6/5 -(-6/5)

= -6/5 + 6/5

= 0

Hence Proved

Answer 2

Steps

If the factor of equation f(x) =ax²+4x+b is (x-3) and (3x-1)

so ,

x-3=0.

x=3

3x-1=0

3x=1

x=1/3

Then ,put x=3 in f(x)

f(3) = a(3)²+4(3)+b

= 9a+12+b

9a+b =-12--------(1)

put x=1/3 in f(x)

f(1/3) = a(1/3)²+4(1/3)+b

=a/9+4/3+b

a/9+b= -4/3

a+9b =-12 -------(2)

9a+b=-12

(2)×9 9a+81b=-108

-80b= 96

b=-96/80

b= -1.2

9a-1.2 =-12

9a=-12+1.2

a=-1.2

Proof

a-b= -1.2+(-1.2)

=0

Hence proved