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POLYNOMIALS

89973 If the zeroes of the quadratic polynomial x $^{1}$ + (a + 1)x + b are 2 and -3, then

1 a = -7,b = -1

2 a = 5, b = -1

3 a = 2, b = -6

4 a = 0, b = -6

Explanation:

a = 0, b = -6
The given quadratic equation is x $^{1}$ + (a + 1)x + b = 0 Since the zeroes of the given equation are 2 and -3. So, $α = 2$ and $β = - 3$ Now, Sum of zeroes $= - \frac{Coefficient of x}{{Coefficient of x}^{2}}$ $\Rightarrow 2 + (- 3) = - \frac{(a + 1)}{1}$ $\Rightarrow - 1 = - a - 1$ $\Rightarrow a = 0$ Product of zeroes $= \frac{Constant of x}{{Coefficient of x}^{2}}$ $\Rightarrow 2 \times (- 3) = \frac{b}{1}$ $\Rightarrow b = - 6$ So, a = 0 and b = -6 Hence, the correct answer is option (d)

POLYNOMIALS

89974 A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is:

1 x

^{1}

- 9

2 x

^{1}

+ 9

3 x

^{1}

+ 3

4 x

^{1}

- 3

Explanation:

x $^{1}$ - 9
Since $α$ and $β$ are the zeros of the quadratic polynomials such that $0 = α + β$ If one of zero is 3 then $α + β = 0$ $3 + β = 0$ $β = 0 - 3$ $β = - 3$ Substituting $β = - 3$ in $α + β = 0$ we get $α - 3 = 0$ $α = 3$ Let S and P denote the sum and product of the zeros of the polynomial respectively then $S = α + β$ $S = 0$ $p = α β$ $p = 3 \times - 3$ $p = - 9$ Hence, the required polynomials is $= (x^{2} - Sx + p)$ $= (x^{2} - 0 x - 9)$ $= x^{2} - 9$ Hence, the correct choice is (a)

POLYNOMIALS

89975 If the polynomial 3x $^{1}$ - 4x $^{1}$ - 17x - k is exactly divisible by x - 3, then the value of k is:

1 6

2 -6

3 5

4 -5

Explanation:

-6
If the polynomial 3x $^{1}$ - 4x $^{1}$ - 17x - k is exactly divisible by x - 3, then p(3) = 0 (By factor theorem) ? 3(3) $^{1}$ - 4(3) $^{1}$ - 17 × 3 - k = 0 ? 81 - 36 - 51 - k = 0 ? -6 - k = 0 ? k = -6

POLYNOMIALS

89976 If $α$ and $β$ are the zeroes of a quadratic polynomial ax $^{1}$ + bx + c, then $α + β =$

1

\frac{c}{a}

2

\frac{b}{a}

3

\frac{- b}{a}

4

\frac{- c}{a}

Explanation:

$\frac{- b}{a}$
If $α$ and $β$ are the zeroes of a quadratic polynomial ax $^{1}$ + bx + c, $∵$ Sum of the zeroes of a quadratic polynomial ax $^{1}$ + bx + c $= \frac{-(Coe f ficient of x)}{{Coe f ficient of x}^{2}} then α + β = \frac{- b}{a}$

POLYNOMIALS

89973 If the zeroes of the quadratic polynomial x $^{1}$ + (a + 1)x + b are 2 and -3, then

1 a = -7,b = -1

2 a = 5, b = -1

3 a = 2, b = -6

4 a = 0, b = -6

Explanation:

a = 0, b = -6
The given quadratic equation is x $^{1}$ + (a + 1)x + b = 0 Since the zeroes of the given equation are 2 and -3. So, $α = 2$ and $β = - 3$ Now, Sum of zeroes $= - \frac{Coefficient of x}{{Coefficient of x}^{2}}$ $\Rightarrow 2 + (- 3) = - \frac{(a + 1)}{1}$ $\Rightarrow - 1 = - a - 1$ $\Rightarrow a = 0$ Product of zeroes $= \frac{Constant of x}{{Coefficient of x}^{2}}$ $\Rightarrow 2 \times (- 3) = \frac{b}{1}$ $\Rightarrow b = - 6$ So, a = 0 and b = -6 Hence, the correct answer is option (d)

POLYNOMIALS

89974 A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is:

1 x

^{1}

- 9

2 x

^{1}

+ 9

3 x

^{1}

+ 3

4 x

^{1}

- 3

Explanation:

x $^{1}$ - 9
Since $α$ and $β$ are the zeros of the quadratic polynomials such that $0 = α + β$ If one of zero is 3 then $α + β = 0$ $3 + β = 0$ $β = 0 - 3$ $β = - 3$ Substituting $β = - 3$ in $α + β = 0$ we get $α - 3 = 0$ $α = 3$ Let S and P denote the sum and product of the zeros of the polynomial respectively then $S = α + β$ $S = 0$ $p = α β$ $p = 3 \times - 3$ $p = - 9$ Hence, the required polynomials is $= (x^{2} - Sx + p)$ $= (x^{2} - 0 x - 9)$ $= x^{2} - 9$ Hence, the correct choice is (a)

POLYNOMIALS

89975 If the polynomial 3x $^{1}$ - 4x $^{1}$ - 17x - k is exactly divisible by x - 3, then the value of k is:

1 6

2 -6

3 5

4 -5

Explanation:

-6
If the polynomial 3x $^{1}$ - 4x $^{1}$ - 17x - k is exactly divisible by x - 3, then p(3) = 0 (By factor theorem) ? 3(3) $^{1}$ - 4(3) $^{1}$ - 17 × 3 - k = 0 ? 81 - 36 - 51 - k = 0 ? -6 - k = 0 ? k = -6

POLYNOMIALS

89976 If $α$ and $β$ are the zeroes of a quadratic polynomial ax $^{1}$ + bx + c, then $α + β =$

1

\frac{c}{a}

2

\frac{b}{a}

3

\frac{- b}{a}

4

\frac{- c}{a}

Explanation:

$\frac{- b}{a}$
If $α$ and $β$ are the zeroes of a quadratic polynomial ax $^{1}$ + bx + c, $∵$ Sum of the zeroes of a quadratic polynomial ax $^{1}$ + bx + c $= \frac{-(Coe f ficient of x)}{{Coe f ficient of x}^{2}} then α + β = \frac{- b}{a}$

POLYNOMIALS

89973 If the zeroes of the quadratic polynomial x $^{1}$ + (a + 1)x + b are 2 and -3, then

1 a = -7,b = -1

2 a = 5, b = -1

3 a = 2, b = -6

4 a = 0, b = -6

Explanation:

a = 0, b = -6
The given quadratic equation is x $^{1}$ + (a + 1)x + b = 0 Since the zeroes of the given equation are 2 and -3. So, $α = 2$ and $β = - 3$ Now, Sum of zeroes $= - \frac{Coefficient of x}{{Coefficient of x}^{2}}$ $\Rightarrow 2 + (- 3) = - \frac{(a + 1)}{1}$ $\Rightarrow - 1 = - a - 1$ $\Rightarrow a = 0$ Product of zeroes $= \frac{Constant of x}{{Coefficient of x}^{2}}$ $\Rightarrow 2 \times (- 3) = \frac{b}{1}$ $\Rightarrow b = - 6$ So, a = 0 and b = -6 Hence, the correct answer is option (d)

POLYNOMIALS

89974 A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is:

1 x

^{1}

- 9

2 x

^{1}

+ 9

3 x

^{1}

+ 3

4 x

^{1}

- 3

Explanation:

x $^{1}$ - 9
Since $α$ and $β$ are the zeros of the quadratic polynomials such that $0 = α + β$ If one of zero is 3 then $α + β = 0$ $3 + β = 0$ $β = 0 - 3$ $β = - 3$ Substituting $β = - 3$ in $α + β = 0$ we get $α - 3 = 0$ $α = 3$ Let S and P denote the sum and product of the zeros of the polynomial respectively then $S = α + β$ $S = 0$ $p = α β$ $p = 3 \times - 3$ $p = - 9$ Hence, the required polynomials is $= (x^{2} - Sx + p)$ $= (x^{2} - 0 x - 9)$ $= x^{2} - 9$ Hence, the correct choice is (a)

POLYNOMIALS

89975 If the polynomial 3x $^{1}$ - 4x $^{1}$ - 17x - k is exactly divisible by x - 3, then the value of k is:

1 6

2 -6

3 5

4 -5

Explanation:

-6
If the polynomial 3x $^{1}$ - 4x $^{1}$ - 17x - k is exactly divisible by x - 3, then p(3) = 0 (By factor theorem) ? 3(3) $^{1}$ - 4(3) $^{1}$ - 17 × 3 - k = 0 ? 81 - 36 - 51 - k = 0 ? -6 - k = 0 ? k = -6

POLYNOMIALS

89976 If $α$ and $β$ are the zeroes of a quadratic polynomial ax $^{1}$ + bx + c, then $α + β =$

1

\frac{c}{a}

2

\frac{b}{a}

3

\frac{- b}{a}

4

\frac{- c}{a}

Explanation:

$\frac{- b}{a}$
If $α$ and $β$ are the zeroes of a quadratic polynomial ax $^{1}$ + bx + c, $∵$ Sum of the zeroes of a quadratic polynomial ax $^{1}$ + bx + c $= \frac{-(Coe f ficient of x)}{{Coe f ficient of x}^{2}} then α + β = \frac{- b}{a}$

POLYNOMIALS

89973 If the zeroes of the quadratic polynomial x $^{1}$ + (a + 1)x + b are 2 and -3, then

1 a = -7,b = -1

2 a = 5, b = -1

3 a = 2, b = -6

4 a = 0, b = -6

Explanation:

a = 0, b = -6
The given quadratic equation is x $^{1}$ + (a + 1)x + b = 0 Since the zeroes of the given equation are 2 and -3. So, $α = 2$ and $β = - 3$ Now, Sum of zeroes $= - \frac{Coefficient of x}{{Coefficient of x}^{2}}$ $\Rightarrow 2 + (- 3) = - \frac{(a + 1)}{1}$ $\Rightarrow - 1 = - a - 1$ $\Rightarrow a = 0$ Product of zeroes $= \frac{Constant of x}{{Coefficient of x}^{2}}$ $\Rightarrow 2 \times (- 3) = \frac{b}{1}$ $\Rightarrow b = - 6$ So, a = 0 and b = -6 Hence, the correct answer is option (d)

POLYNOMIALS

89974 A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is:

1 x

^{1}

- 9

2 x

^{1}

+ 9

3 x

^{1}

+ 3

4 x

^{1}

- 3

Explanation:

x $^{1}$ - 9
Since $α$ and $β$ are the zeros of the quadratic polynomials such that $0 = α + β$ If one of zero is 3 then $α + β = 0$ $3 + β = 0$ $β = 0 - 3$ $β = - 3$ Substituting $β = - 3$ in $α + β = 0$ we get $α - 3 = 0$ $α = 3$ Let S and P denote the sum and product of the zeros of the polynomial respectively then $S = α + β$ $S = 0$ $p = α β$ $p = 3 \times - 3$ $p = - 9$ Hence, the required polynomials is $= (x^{2} - Sx + p)$ $= (x^{2} - 0 x - 9)$ $= x^{2} - 9$ Hence, the correct choice is (a)

POLYNOMIALS

89975 If the polynomial 3x $^{1}$ - 4x $^{1}$ - 17x - k is exactly divisible by x - 3, then the value of k is:

1 6

2 -6

3 5

4 -5

Explanation:

-6
If the polynomial 3x $^{1}$ - 4x $^{1}$ - 17x - k is exactly divisible by x - 3, then p(3) = 0 (By factor theorem) ? 3(3) $^{1}$ - 4(3) $^{1}$ - 17 × 3 - k = 0 ? 81 - 36 - 51 - k = 0 ? -6 - k = 0 ? k = -6

POLYNOMIALS

89976 If $α$ and $β$ are the zeroes of a quadratic polynomial ax $^{1}$ + bx + c, then $α + β =$

1

\frac{c}{a}

2

\frac{b}{a}

3

\frac{- b}{a}

4

\frac{- c}{a}

Explanation:

$\frac{- b}{a}$
If $α$ and $β$ are the zeroes of a quadratic polynomial ax $^{1}$ + bx + c, $∵$ Sum of the zeroes of a quadratic polynomial ax $^{1}$ + bx + c $= \frac{-(Coe f ficient of x)}{{Coe f ficient of x}^{2}} then α + β = \frac{- b}{a}$