the expression x^4+x^3-ax+11 and x^3+2x^2-3 leaves the remainder when divided by (x-2) find the value of a
Answers
Answer:
Let us take x–2=0
Then, x=2
Given, f(x)=2x
3
+ax
2
–11x+b
Now, substitute the value of x in f(x),
f(2)=2(2)
3
+a(2)
2
–11(2)+b
=16+4a–22+b
=–6+4a+b
Given, remainder is 0.
So,−6+4a+b=0
4a+b=6 … [equation (i)]
Now, consider (x – 3)
Assume x–3=0
Then, x=3
Given,f(x)=2x
3
+ax
2
–11x+b
Now, substitute the value of x in f(x),
f(2)=2(3)
3
+a(3)
2
–11(3)+b
=54+9a–33+b
=21+9a+b
Given, remainder is 42.
So, 21+9a+b=42
9a+b=42–21
9a+b=21 … [equation (ii)]
Now, subtracting equation (i) from equation (ii) we get,
(9a+b)–(4a+b)=21–6
9a+b–4a–b=15
5a=15
a=15/5
a=3
Consider the equation (i) to find out ‘b’.
4a+b=6
4(3)+b=6
12+b=6
b=6–12
b=−6
Then, by substituting the value of a and bf(x)=2x
3
+3x
2
–11x–6
Given that remainder is 0 for, (x–2) is a factor of f(x).
So, dividing f(x) by (x–2)
Therefore, 2x
3
+3x
2
–11x–6=(x–2)(2x
2
+7x+3)
=(x–2)(2x
2
+6x+x+3)
=(x–2)(2x
2
+6x+x+3)
=(x–2)(2x(x+3)+1(x+3))
=(x–2)(x+3)(2x+1)