Question

I need the answer for 3 question in first Roman letter n also 4,5 in second roman

Answer 1

I
3sol)
Let given polynomial be
f(x)=x³-3x²+kx+42
and (x+3) is a factor of f(x)
Therefore -3 is a zero of f(x)
Now we have,
f(x)=0

f(-3)=0

(-3)³-3(-3)²+k(-3)+42=0

-27-27-3k+42=0

-54-3k+42=0

-12=3k

k=-4

Hence value of k=-4

II
4sol)
Let given polynomial be,
f(x)=2x²+7x-4
'a' is said to be zero of f(x) when f(a)=0.
Then we have,
f(x)=0

f(1/2)=2(1/2)²+7(1/2)-4

f(1/2)=2/4+7/2-4

f(1/2)=1/2+7/2-4

f(1/2)=0

As f(1/2)=0
Therefore 1/2 is a zero of given polynomial f(x).


5sol)
Given that,
Product of zeroes=-13/5
Sum of zeroes=-3/5
We know that quadratic polynomial is in the form of,
=k{x²-(sum of the zeroes)x+product of zeroes}
By putting given values we get,
=k{x²-(-3/5)x-13/5}

=k{x²+3x/5-13/5}
Putting k=5 we get,
=5{x²+3x/5-13/5}

=5x²+3x-13

Hence required polynomial is 5x²+3x-13