Question

If the polynomials az³ + 4z² + 3z -4 and z³ - 4z + a leave the same remainder when divided by z- 3, find the value of a​

Answer 1

Given :

$\longrightarrow{Cubic\: expression\:is\:a{z}^{3}+4{z}^{2}+3z-4}$

$\longrightarrow{Cubic\: expression\:is\:{z}^{3}-4z+a }$

Required to find :

$\large{\longrightarrow{\boxed{Value\:of\:a}}}{\longleftarrow}$

Mentioned hints :

☞ When both expression are divided they give us the same remainder .

Explanation :

In the question we were given with two expressions .

The two expressions are cubic expressions ;

So, let's add their functions .

Then , the 1st cubic expressions becomes ;

$\longrightarrow{p(z)\:=\:a{z}^{3}+4{z}^{2}+3z-4}$

Similarly, 2nd cubic expression becomes ;

$\longrightarrow{q(z)\:=\:{z}^{3}-4z+a}$

Now, we have to equal the value of z - 3 to 0

Hence, we can find the value of z .

Now, we have to substitute the z value in place of z in p(z) &. q(z) .

First we have to to substitute this value in p(z) and we solve the question hence we will be left with some remainder .

Now, we have to the substitute the z value in q(z) and we have to add the above remainder to the R.H.S part of the expression . (this is because the remainder is same in both cases ).

The R.H.S part of the expression actually represents the remainder .

On solving further we will be left with the value of " a " .

So, now let's crack the above question .

Solution :

Consider the first expression ;

$\longrightarrow{p(z)\:=\:a{z}^{3}+4{z}^{2}+3z-4}$

As it is given that ( z - 3 ) when divided leaves remainder .

So, Let ,

$\Rightarrow{z - 3 = 0}$

$\rightarrow{\boxed{z = 3}}$

Hence,

Substitute this value in place of z in expression 1 i.e. p(z) .

$\longrightarrow{p(3)\:=\:a{(3)}^{3}+4{(3)}^{2}+3(3)-4}$

$\longrightarrow{=\:a(27)+4(9)+9-4}$

$\longrightarrow{=\:27a + 36 + 9 - 4 }$

$\longrightarrow{=\:27a + 45 - 4}$

$\longrightarrow{=\:27a + 41 }$

Hence, on dividing the p(z) the remainder is 27a + 41 .

The remainder is same in both cases ,

Hence,

substitute the z value in q(z) .

$\longrightarrow{q(3) = }$

$\longrightarrow{{(3)}^{3}-4(3)+a = 27a + 41 }$

$\longrightarrow{27 - 12 + a = 27a + 41}$

$\longrightarrow{27 - 12 + a - 27a - 41 = 0 }$

$\longrightarrow{- 26a - 12 - 41 + 27 = 0 }$

$\longrightarrow{ - 26a - 53 + 27 = 0}$

$\longrightarrow{ - 26a - 26 = 0 }$

$\longrightarrow{ - 26a = 26 }$

$\longrightarrow{ a = \frac{26}{-26}}$

$\longrightarrow{ a = \cancel {\frac{26}{-26}}}$

$\implies{ a = -1}$

$\large{\longrightarrow{\boxed{\Rightarrow{\boxed{\therefore{Value\:of\:a\:is\:-1}}}}}}$