answer the 12 th one plss........
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Let p(x)=(x+1)⁷+(2x+k)³
∵p(x) is exactly divisible by (x+2)
∴(x+2) is a factor of p(x)
∴x+2=0
⇒x=-2
∴p(-2)=(-2+1)⁷+{2(-2)+k}³
⇒(-1)⁷+(k-4)³=0[∵(x+2) is a factor of p(x)]
⇒-1+{k³-3.k².4+3.k.(4)²-4³}=0
⇒-1+k³-12k²+48k-64=0
⇒k³-12k²+48k-65=0 ........(1)
now the factors of constant term 65 are ±5 & ±13
using k =5 in L.H.S of (1)
(5)³-12(5)²+48(5)-65
=125-300+240-65
=0
∴k=5 is a root
∴k=5 satisfies equation (1)
∴k=5 is a correct value
∴option(3) is correct.
∵p(x) is exactly divisible by (x+2)
∴(x+2) is a factor of p(x)
∴x+2=0
⇒x=-2
∴p(-2)=(-2+1)⁷+{2(-2)+k}³
⇒(-1)⁷+(k-4)³=0[∵(x+2) is a factor of p(x)]
⇒-1+{k³-3.k².4+3.k.(4)²-4³}=0
⇒-1+k³-12k²+48k-64=0
⇒k³-12k²+48k-65=0 ........(1)
now the factors of constant term 65 are ±5 & ±13
using k =5 in L.H.S of (1)
(5)³-12(5)²+48(5)-65
=125-300+240-65
=0
∴k=5 is a root
∴k=5 satisfies equation (1)
∴k=5 is a correct value
∴option(3) is correct.
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