@ If (2-1) (a) and (2x+3
and (2x+3) are in ne
when the value of
X
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Given polynomial p(x)=ax
3
+3x
2
−13 and 2x
2
−5x+a is divided by x-2 get same remainder
Then x-2=0 or x=2
p(x)=ax
3
+3x
2
−13
Replace x by 2 we get
p(2)=a(2)
3
+3(2)
2
−13
⇒p(2)=8a+12−13
⇒p(2)=8a−1
2x
2
−5x+a
Replace x by 2 we get
q(2)=2(2)
3
−5(2)+a
⇒q(2)=16−10+a
⇒q(2)=6+a
Given remainder is same
∴8a−1=6+a
⇒8a−1+1−a=6+a+1−a
⇒7a=7
a=1
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