if both x-4 and x-1/4 are factors of ax^2+5x+b , show that a=b
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let p(x)=ax²+5x+b,
since (x-4) is a factor of polynomial,
then
p(4)=0,
a×4²+5×4+b=0,
16a+20+b=0,.......eq(1),
since (x-1/4) is a factor of p(x), then
p(1/4) =0,
a×(1/4)²+5×1/4+b=0,
a/16 + 5/4 +b=0,
a+20 + 16b=0,.........eq(2),
on solving these two equations we will get
a=b
since (x-4) is a factor of polynomial,
then
p(4)=0,
a×4²+5×4+b=0,
16a+20+b=0,.......eq(1),
since (x-1/4) is a factor of p(x), then
p(1/4) =0,
a×(1/4)²+5×1/4+b=0,
a/16 + 5/4 +b=0,
a+20 + 16b=0,.........eq(2),
on solving these two equations we will get
a=b
vasukishesh:
in eq 2 , how did a/16+5/4+b get converted to a+20+4b ??
ok
shouldn't it be 16b then...?
ok
ty
but again , shouldn't the statement have 16 as denominator ....(a+20+4b)÷16??
ok and ty once again
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