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Published on: 08/10/2019
Polynomials
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1.
Write the cubic polynomial, whose zeroes are \(2-2\sqrt { 5 } ,2+2\sqrt { 5 }\) and 1, respectively.
2.
Draw the graph of the polynomial -x2+4x-4 and find the zeroes of the polynomial.
3.
If the polynomial f(x)=3x4-9x3+x2+15x+k is completely divisible by 3x2-5, then find the value of k and hence the other two zeroes of the polynomial.
4.
A polynomial g(x) of degree zero is added to the polynomial 2x3+5x2-14x+10, so that it becomes exactly divisible by 2x-3. Find g(x).
5.
If the polynomial 6x4+8x3+17x2+21x+7 is divided by another polynomial 3x2+4x+1, the remainder comes out to be ax+b, then find the values of a and b.
6.
On dividing polynomial p(x) by 3x+1, the quotient is 2x-3 and the remainder is -2. Find p(x).
7.
If α and β are zeroes of the quadratic polynomial p(x)=6x2+x-1, then find the value of \(\frac { \alpha }{ \beta } +\frac { \alpha }{ \alpha } +2\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } \right) +3\alpha \beta \)
8.
Find the zeroes of quadratic polynomial y2+92y+1920.
9.
If α and β are the zeroes of the quadratic polynomial f(x)=3x2-5x-2, then evaluate α3+β3.
10.
If zeroes α and β of a polynomial x2-7x+k are such that α-β=1, then find the value of k.
1.
x3-5x2-12x+16
2.
2, 2
3.
On dividing f(x) by 3x2-5, we get
Quotient=x2-3x+2 and remainder =k+10.
Since, f(x) is exactly divisible by 3x2-5, so remainder =0.
⇒ k+10=0⇒ k-10
Now, for other zeroes, put x2-3x+2=0
⇒ (x2)(x1)=0⇒x=1,2
4.
Let g(x)=k, then 2x 3 +5x 2 -14x+10+k will be exactly by 2x-3.
On dividing 2x 3 +5x 2 -14x+10+k by 2x-3, we get Quotient x2+4x-1 and remainder 7+k.
Since, 2x 3 +5x 2 -14x+10+k is exactly divisible by 2x-3, so remainder=0
⇒ 7+k=0⇒k=-7
5.
On dividing 6x4+8x3+17x2+21x+7 by 3x2+4x+1,
We get
Quotient=2x2+5 and remainder=x+2
But, its is given that the remainder is ax+b.
So, ax+b=x+2
On comparing the coefficients of x and constant terms, we get a=1 and b=2
6.
By division algorithm, p(x)=g(x).q(x)+r(x)=6x2-7x-5
7.
\(\frac { \alpha }{ \beta } +\frac { \alpha }{ \alpha } +2\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } \right) +3\alpha \beta \)
\(=\frac { \alpha ^{ 2 }+\beta ^{ 2 } }{ \alpha \beta } +2\left( \frac { \alpha +\beta }{ \alpha \beta } \right) +3\alpha \beta =-\frac { 2 }{ 3 } \)
8.
Let p(y)=y2+92y+1920=(y+32)(y+60)
Now, for zeroes of p(y), put p(y)=0
Zeroes y=-32, -60
9.
α3+β3=(α3+β3)-3αβ(α+β) \(\frac { 215 }{ 27 } \)
10.
α+β=7 ...(i)
α-β=1 ....(ii)
On solving Eqs. (i) and (ii), we get α=4 and β=3
Now, k=αβ=12
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CBSE 10th Standard CBSE Subjects
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