If (x−3) and (x+4)(x−3) and (x+4) are factors of polynomial P(x)P(x) , then
Answers
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Step-by-step explanation:
Let us assume x+3=0
Then, x=−3
Given, f(x)=x
3
+ax
2
–bx+24
Now, substitute the value of x in f(x),
f(−3)=(−3)
3
+a(−3)
2
–b(−3)+24
=−27+9a+3b+24
=9a+3b–3
Dividing all terms by 3 we get,
=3a+b–1
From the question, (x+3) is a factor of x
3
+ax
2
–bx+24.
Therefore, remainder is 0.
f(x)=0
3a+b–1=0
3a+b=1 … [equation (i)]
Now, assume x–4=0
Then, x=4
Given, f(x)=x
3
+ax
2
–bx+24
Now, substitute the value of x in f(x),
f(4)=43+a(4)
2
–b(4)+24
=64+16a–4b+24
=88+16a–4b
Dividing all terms by 4 we get,
=22+4a–b
From the question, (x–4) is a factor of x
3
+ax
2
–bx+24.
Therefore, remainder is 0.
f(x)=0
22+4a–b=0
4a–b=–22 … [equation (ii)]
Now, adding both equation (i) and equation (ii) we get,
(3a+b)+(4a–b)=1–22
3a+b+4a–b=–21
7a=–21
a=−21/7
a=−3
Consider the equation (i) to find out ‘b’.
3a+b=1
3(−3)+b=1
−9+b=1
b=1+9
b=10
Therefore, value of a=−3 and b=10.
Then, by substituting the value of a and bf(x)=x
3
–3x
2
–10x+24
(x+3)(x–4)
=x(x–4)+3(x–4)
=x2–4x+3x–12
=x
2
–x–12
Dividing f(x) by x
2
–x–12 we get,
Therefore, x
3
–3x2–10x+24=(x2–x–12)(x–2)
=(x+3)(x–4)(x–2)