x2+x−12
=x2+4x−3x−12
=x(x+4)−3(x+4)=(x−3)(x+4)
So zeroes are 3 and -4
for f(x)=x2+7x+7
we get
a+b=−7
ab=7
Now
a−1+b−1−2ab
=a+b−2a2b2ab
=−7−987=−15
Given p(x)=x3−ax2+6−a
By remainder theorem
p(a) = 6-a
for f(x)=x2−5x+k
we get
p+q=5
pq=k
Now
p−q=1
(p−q)2=1
(p+q)2−4pq=1
25−4k=1
k=6
Two zeroes = 0, 0
Let the third zero be k.
The, using relation between zeroes and coefficient of polynomial, we have:
k+0+0=−ba
Third zero = k = -b/a
Let f(x)=x2+(a+1)x+b
Then
2−3=−(a+1)ora=0
−6=b
So a=0 and b=6
k1k2k3=−ConstanttermCoefficientofx3
Hence
k1k2k3=−c
Now k1=−1
So,
k2k3=c
Let f(x)=x3−6x2+8x
Method -1
=x(x2−6x+8)=x(x−2)(x−4)
So 0,2,4 are zeroes of the polynomial. or a=2 and b=2 or -2
Method -2
k1+k2+k3=−Coefficientofx2Coefficientofx3
a−b+a+a+b=6 or a=2
Now k1k2k3=−ConstanttermCoefficientofx3
(2−b)2(2+b)=0 or b=+2 or -2
f(x)=2x2−7x+3
=2x2−x−6x+3=x(2x−1)−3(2x−1)=(x−3)(2x−1)
So zeroes are 3 and 1/2
Now
a2+b2
=9+14=374
4x2+12x+9
=4x2+6x+6x+9
=2x(2x+3)+3(2x+3)=(2x+3)2
So p=-3/2 and q=-3/2
So, p−1=−52 and q−1=−52
So quadratic polynomial will be
(x+52)2
or
4x2+20x+25
Let f(x)=2x3+x2−5x+2
For Verification of zeroes, we can simply substitute in the polynomial and verify
f(1/2) = 2 (\frac {1}{2}}^3 + (\frac {1}{2})^2 -5 \frac {1}{2} + 2 = 0
f(1)=2(1)3+(1)2−5(1)+2=0
f(2)=2(2)3+(2)2−5(2)+2=0
The relationship between the zeroes and coefficient is given by
i. k1+k2+k3=−Coefficientofx2Coefficientofx3
or
12+1−2=−12
−12=−12
Verified
ii. k1k2k3=−ConstanttermCoefficientofx3
12×1×(−2)=−1
-1=-1}
Verified
iii.k_1 k_2 + k_2 k_3 + k_1 k_3 = \frac {Coefficient \; of \; x}{Coefficient \; of \; x^3} \frac {1}{2} -2 -1 = \frac {-5}{2} \frac {-5}{2}= \frac {-5}{2}$
Verified
for f(x)=x2−(k+6)x+2(2k−1)
We get,
p+q=k+6
pq=2(2k−1)
Now
2(p+q)=pq
Therefore,
2(k+6)=2(2k−1)
or k+6=2k−1
or k=7
for f(x)=ax2−5x+c
we get
m+n=5a
mn=ca
Given m+n=m×n=10
Therefore,
5/a=10
a=1/2
c/a=10
or c=5
k1+k2=√2
k1k2=−32
Now the quadratic polynomial is given by
x2−(k1+k2)x+k1k2
=x2−√2x−32
or
=2x2−2√2x−3
Now we need to find the zeroes of this polynomial
2x2−2√2x−3
=2x2−3√2x+√2x−3
=√2x(√2x−3)+1(√2x−3)
=(√2x+1)(√2x−3)
or roots are -\frac {1}{\sqrt {2}} , \frac {3}{\sqrt {2}
k1+k2+k3=−Coefficientofx2Coefficientofx3
a+a+b+a+2b=6 or a+b=2 or a=2−b
Now
k1k2k3=−ConstanttermCoefficientofx3
a(a+b)(a+2b)=−10
2(2−b)(2+b)=−10
(4−b2)=−5
or b= -3 or 3
So a= 5 or -1
The zeroes with a=5 ,b=-3 can be expressed as 5 ,2,-1
The zeroes with a=-1 ,b=3 can be expressed as -1 ,2,5
Here is the division
So f(x)=(2x2−1)g(x)
Let a be one zero ,then another will be 2a
Now
a+2a=52 or a= 5/6
Also
a×2a=−(2k+1)2
or
259=−(2k+1)
2k=−349
or
k= -17/9
for f(x)=x2−px+q
We get
m+n=p
mn=q
a. m2+n2=(m+n)2−2mn=p2−q
b. m−1+n−1=m+nmn=pq
Let a-b,a ,a+b are the zeroes
Now
k1+k2+k3=−Coefficientofx2Coefficientofx3
a−b+a+a+b=12 or a=4
Also
k1k2+k2k3+k1k3=CoefficientofxCoefficientofx3
(a−d)a+a(a+d)+(a+d)(a−d)=39
3a2−d2=39
Substituting the value of a from above
3(4)2−d=39
d=±9
hence, series are 1 , 4 , 7 or 7, 4 , 1
now,
k1k2k3=−ConstanttermCoefficientofx3
7×4×1=−k
k=-28
Dividend = Divisor X Quotient + Remainder
Dividend - Remainder = Divisor X Quotient
Dividend - Remainder is always divisible by the divisor.
Now, it is given that f(x) when divided by x2 -2x + k leaves (x + a) as remainder.
So f(x) -(x+a) is divided by x2 -2x + k
x4 -6x3 + 16x2 - 25x + 10 -(x+a) =x4 -6x3 + 16x2 - 26x + 10 -a
Now doing the division
Now remainder should be zero
(-10 + 2k)x + (10 - a - 8k + k2) = 0
-10 + 2k = 0 or k=5
Now (10 -a - 8k + k2) = 0
or 10 - a - 8 (5) + 52 = 0
or - a - 5 = 0
a =-5
√2 , -√2, 2,1
k= -3
The other root is 3/2
61x - 65
4q+10+r=0 -(1)
q/4 +5/2 +r=0 or q+10+4r=0 -(2)
Subtracting 1 from 2
3q-3r=0
q=r
For x2 + 2x + k is a factor of polynomial 2x4 + x3 - 14x2 + 5x + 6, it should be able to divide the polynomial without any remainder
Comparing the coefficient of x we get.
21+7k=0 or k=-3
So x2 + 2x + k becomes x2 + 2x -3 = (x-1)(x+3)
Now
2x4 + x3 - 14x2 + 5x + 6= (x2 + 2x -3)(2x2-3x-8+2k)
=(x2 + 2x -3)(2x2-3x-2)
=(x-1)(x+3)(x-2)(2x+1)
or x= 1,-3,2,=-1/2
x2 - x +1
By cubic polynomial equation, we have
a+b+c=-p
ab+bc+ac=q
abc=-2
Now given ab+1=0 or ab=-1
So abc=-2
(-1)c=-2 or c=2
2p + q + 5
=-2(a+b+c) +(ab+bc+ac) +5
=-2(a+b+2) +[-1+2(a+b)] +5
=-4 -1+5=0
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