Can anyone tell me the ans for Question no 8 and 12 plz it's urgent tommorow is my exam so plz...ans
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I left those portions. After exam
I dont remeber...
Is it needed now? I will try to ans tmrw
Its not in google
I think we must put -4 in plce of x
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8. p(x)=x2-2x-(7p+4)
One zeros is -4
Let x be -4
x+4 is the factor of p(x)
Let x be -4
p(x)=x2-2x-7p-4
p(-4)=(-4)2-2(-4)-7p-4
=16+8-7p-4
=24-7p-4
=20-7p
-7p=-20
The value of p is,
p=20/7
The required polynomial is,
x2-2x-(7p+4)
x2-2x-7×20/7-4
x2-2x-20-4
=x2-2x-24
Let a be -4
Sum of zeros = -b/a
a+b=-(-2)/1
=2
-4+b=2
b=2+4
b=6
So, the other zeros is 6
I hope it's helps u
If u like my answer please mark me as brainliest
One zeros is -4
Let x be -4
x+4 is the factor of p(x)
Let x be -4
p(x)=x2-2x-7p-4
p(-4)=(-4)2-2(-4)-7p-4
=16+8-7p-4
=24-7p-4
=20-7p
-7p=-20
The value of p is,
p=20/7
The required polynomial is,
x2-2x-(7p+4)
x2-2x-7×20/7-4
x2-2x-20-4
=x2-2x-24
Let a be -4
Sum of zeros = -b/a
a+b=-(-2)/1
=2
-4+b=2
b=2+4
b=6
So, the other zeros is 6
I hope it's helps u
If u like my answer please mark me as brainliest
Welcome and please add me in brainliest
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