Let we have an equation $x^2+4x+2$ and we want to divide it with $x-4$. By remainder theorem it gives the value of $34$. But when we put the value of $x=6$ in equations it gives zero remainder. And quotient is $31$. How's that? Where I am mistaking?
Asked
Active
Viewed 36 times
1
-
How do you get zero remainder with $x=6$? – Angina Seng Oct 09 '17 at 16:36
-
The first polynomial gives the value 62 when x=6 and second give 2. When 62 is divided by 2 remainder is 0 – Bilal Sheikh Oct 09 '17 at 16:39
-
$f(x)=x^2+4x+2=(x+8)(x-4)+34$ is correct. As for "when we put in the value of $x=6$ it gives zero remainder and quotient 31", that sounds incorrect. $f(x)=x^2+4x+2=(x+10)(x-6)+62$. I don't see where you got zero or 31 or why you might choose to talk about dividing by two. – JMoravitz Oct 09 '17 at 16:39
2 Answers
2
You are getting remainder 34. This is because you haven't placed value of x yet. If you place value of x, you will see that divisor is (6-4) = 2. Now obviously you can't get remainder of 34 when divisor is 2.(remainder can't be greater than divisor) You get remainder 0.
The equation you solved earlier was general case. So if x had been any other number greater than 38(so that divisor is greater than 34), then remainder would have been 34. Otherwise you would have to divide remaining term with divisor
Sagar Chand
- 1,682
1
$x^2 + 4x +2=(x+8)(x-4)+34$. I think you're mistaking what dividing a polynomial means
Francisco José Letterio
- 2,146
-
1+1 ... When you substitute $6$ for $x$ here both sides are the same $62$. – Ethan Bolker Oct 09 '17 at 16:56