Question

The polynomials (ax^3 + 3x - 3) and (2x - 5x + a) when die
(x - 4) leave the same remainder. Find the value of a.​

Answer 1

Answer:

1

Step-by-step explanation:

Given=》

ax³+3x²-3 and 2x³-5x+a

When divided by (x-4) leave the remainder same

To find =》

value of a

Solution =》

p(x) =ax³+3x²-3

g(x) = x-4

x-4=0

x=4

putting x=4 in p(x)

a×4³+3×4²-3=0

64a+3×16-3=0

64a+48-3=0

64a+45=0--------------(1)

And

p(x) = 2x³-5x+a

g(x) = x-4

x-4=0

x=4

putting x=4 in p(x)

2×4³-5×4+a=0

2×64-20+a=0

128-20+a=0

a+108=0-----------(2)

Subtract eq 2 from 1 , we get

64a + 45 = 0

a + 108 = 0

- -

————————

63a-63=0

63a=63

a=63/63

a=1

So, the value of a is 1

Answer 2

Solution :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

The polynomials (ax³ + 3x² - 3) and (2x³ -5x + a) when divided by (x-4) leaves the same remainder.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The value of a.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

We have;

• p(x) = 2x³ + 3x² - 3
• q(x) = 2x³ - 5x + a

A/q

x - 4 = 0

x = 4

p(4) = q(4)

$\mapsto\sf{a(4)^{3} +3(4)^{2} -3=2(4)^{3} -5(4)+a}\\\\\mapsto\sf{a*64+3*16-3=2*64-20+a}\\\\\mapsto\sf{64a+48-3=128-20+a}\\\\\mapsto\sf{64a+45=108+a}\\\\\mapsto\sf{64a-a=108-45}\\\\\mapsto\sf{63a=63}\\\\\mapsto\sf{a=\cancel{\dfrac{63}{63} }}\\\\\mapsto\sf{\pink{a=1}}$

Thus;

The value of a is 1 .