if p(x) = az + 4z + 3z - 4 and q(s) leave the same remainders when divided by z - 3 find the value of a
Answers
Answer:
Cubicexpressionisaz
3
+4z
2
+3z−4
\longrightarrow{Cubic\: expression\:is\:{z}^{3}-4z+a }⟶Cubicexpressionisz
3
−4z+a
Required to find :
\large{\longrightarrow{\boxed{Value\:of\:a}}}{\longleftarrow}⟶
Valueofa
⟵
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}⟶p(z)=az
3
+4z
2
+3z−4
Similarly, 2nd cubic expression becomes ;
\longrightarrow{q(z)\:=\:{z}^{3}-4z+a}⟶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}⟶p(z)=az
3
+4z
2
+3z−4
As it is given that ( z - 3 ) when divided leaves remainder .
So, Let ,
\Rightarrow{z - 3 = 0}⇒z−3=0
\rightarrow{\boxed{z = 3}}→
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}⟶p(3)=a(3)
3
+4(3)
2
+3(3)−4
\longrightarrow{=\:a(27)+4(9)+9-4}⟶=a(27)+4(9)+9−4
\longrightarrow{=\:27a + 36 + 9 - 4 }⟶=27a+36+9−4
\longrightarrow{=\:27a + 45 - 4}⟶=27a+45−4
\longrightarrow{=\:27a + 41 }⟶=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) = }⟶q(3)=
\longrightarrow{{(3)}^{3}-4(3)+a = 27a + 41 }⟶(3)
3
−4(3)+a=27a+41
\longrightarrow{27 - 12 + a = 27a + 41}⟶27−12+a=27a+41
\longrightarrow{27 - 12 + a - 27a - 41 = 0 }⟶27−12+a−27a−41=0
\longrightarrow{- 26a - 12 - 41 + 27 = 0 }⟶−26a−12−41+27=0
\longrightarrow{ - 26a - 53 + 27 = 0}⟶−26a−53+27=0
\longrightarrow{ - 26a - 26 = 0 }⟶−26a−26=0
\longrightarrow{ - 26a = 26 }⟶−26a=26
\longrightarrow{ a = \frac{26}{-26}}⟶a=
−26
26
\longrightarrow{ a = \cancel {\frac{26}{-26}}}⟶a=
−26
26
\implies{ a = -1}⟹a=−1
\large{\longrightarrow{\boxed{\Rightarrow{\boxed{\therefore{Value\:of\:a\:is\:-1}}}}}}⟶
⇒
∴Valueofais−1