Answer :
Sure, I'd be happy to help with these questions.
### Question 14:
To find a quadratic polynomial with one zero as [tex]\( \sqrt{5} \)[/tex] and the product of zeroes as [tex]\( -2\sqrt{5} \)[/tex]:
1. Let the other zero be [tex]\( \alpha \)[/tex]. You know:
[tex]\[ \text{Product of zeroes} = \sqrt{5} \times \alpha = -2\sqrt{5} \][/tex]
2. Solve for [tex]\( \alpha \)[/tex]:
[tex]\[ \alpha = \frac{-2\sqrt{5}}{\sqrt{5}} = -2 \][/tex]
3. Sum of zeroes:
[tex]\[ \text{Sum of zeroes} = \sqrt{5} + (-2) = \sqrt{5} - 2 \][/tex]
4. Form the quadratic polynomial using the sum and product of zeroes:
A quadratic polynomial with zeroes [tex]\( p \)[/tex] and [tex]\( q \)[/tex] is given by:
[tex]\[ x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \][/tex]
Substituting the values, we get:
[tex]\[ x^2 - (\sqrt{5} - 2)x - 2\sqrt{5} \][/tex]
So, the quadratic polynomial is:
[tex]\[ x^2 - (\sqrt{5} - 2)x - 2\sqrt{5} \][/tex]
### Question 15:
To show that if [tex]\(1\)[/tex] and [tex]\(-1\)[/tex] are zeroes of [tex]\( a x^4 - b x^3 + c x^2 + d x + e = 0 \)[/tex] then [tex]\(a + c + e = 0\)[/tex] and [tex]\(-b + d = 0\)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into the polynomial:
[tex]\[ a(1)^4 - b(1)^3 + c(1)^2 + d(1) + e = 0 \][/tex]
Simplifying, we get:
[tex]\[ a - b + c + d + e = 0 \quad \text{(Equation 1)} \][/tex]
2. Substitute [tex]\( x = -1 \)[/tex] into the polynomial:
[tex]\[ a(-1)^4 - b(-1)^3 + c(-1)^2 + d(-1) + e = 0 \][/tex]
Simplifying, we get:
[tex]\[ a + b + c - d + e = 0 \quad \text{(Equation 2)} \][/tex]
3. Add Equation 1 and Equation 2:
[tex]\[ (a - b + c + d + e) + (a + b + c - d + e) = 0 + 0 \][/tex]
[tex]\[ 2a + 2c + 2e = 0 \][/tex]
Simplifying, we get:
[tex]\[ a + c + e = 0 \][/tex]
4. Subtract Equation 2 from Equation 1:
[tex]\[ (a - b + c + d + e) - (a + b + c - d + e) = 0 - 0 \][/tex]
[tex]\[ -2b + 2d = 0 \][/tex]
Simplifying, we get:
[tex]\[ -b + d = 0 \][/tex]
Thus, we have proven that [tex]\( a + c + e = 0 \)[/tex] and [tex]\( -b + d = 0 \)[/tex].
### Question 16:
To find the value of [tex]\( b \)[/tex] for which [tex]\( (2x + 3) \)[/tex] is a factor of [tex]\( 2x^3 + 9x^2 - x - b \)[/tex]:
1. Factor Theorem: If [tex]\( (2x + 3) \)[/tex] is a factor, then [tex]\( 2x + 3 = 0 \implies x = -\frac{3}{2} \)[/tex] will satisfy the polynomial.
2. Substitute [tex]\( x = -\frac{3}{2} \)[/tex] into the polynomial:
[tex]\[ 2\left(-\frac{3}{2}\right)^3 + 9\left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right) - b = 0 \][/tex]
3. Calculate each term:
[tex]\[ 2\left(-\frac{27}{8}\right) + 9\left(\frac{9}{4}\right) + \frac{3}{2} - b = 0 \][/tex]
[tex]\[ -\frac{54}{8} + \frac{81}{4} + \frac{3}{2} - b = 0 \][/tex]
[tex]\[ -\frac{27}{4} + \frac{81}{4} + \frac{3}{2} - b = 0 \][/tex]
[tex]\[ \frac{54}{4} + \frac{3}{2} - b = 0 \][/tex]
[tex]\[ \frac{27}{2} + \frac{3}{2} - b = 0 \][/tex]
[tex]\[ \frac{30}{2} - b = 0 \][/tex]
[tex]\[ 15 - b = 0 \][/tex]
[tex]\[ b = 15 \][/tex]
Thus, the value of [tex]\( b \)[/tex] for which [tex]\( 2x + 3 \)[/tex] is a factor of [tex]\( 2x^3 + 9x^2 - x - b \)[/tex] is [tex]\( b = 15 \)[/tex].
### Question 14:
To find a quadratic polynomial with one zero as [tex]\( \sqrt{5} \)[/tex] and the product of zeroes as [tex]\( -2\sqrt{5} \)[/tex]:
1. Let the other zero be [tex]\( \alpha \)[/tex]. You know:
[tex]\[ \text{Product of zeroes} = \sqrt{5} \times \alpha = -2\sqrt{5} \][/tex]
2. Solve for [tex]\( \alpha \)[/tex]:
[tex]\[ \alpha = \frac{-2\sqrt{5}}{\sqrt{5}} = -2 \][/tex]
3. Sum of zeroes:
[tex]\[ \text{Sum of zeroes} = \sqrt{5} + (-2) = \sqrt{5} - 2 \][/tex]
4. Form the quadratic polynomial using the sum and product of zeroes:
A quadratic polynomial with zeroes [tex]\( p \)[/tex] and [tex]\( q \)[/tex] is given by:
[tex]\[ x^2 - (\text{sum of zeroes})x + (\text{product of zeroes}) \][/tex]
Substituting the values, we get:
[tex]\[ x^2 - (\sqrt{5} - 2)x - 2\sqrt{5} \][/tex]
So, the quadratic polynomial is:
[tex]\[ x^2 - (\sqrt{5} - 2)x - 2\sqrt{5} \][/tex]
### Question 15:
To show that if [tex]\(1\)[/tex] and [tex]\(-1\)[/tex] are zeroes of [tex]\( a x^4 - b x^3 + c x^2 + d x + e = 0 \)[/tex] then [tex]\(a + c + e = 0\)[/tex] and [tex]\(-b + d = 0\)[/tex]:
1. Substitute [tex]\( x = 1 \)[/tex] into the polynomial:
[tex]\[ a(1)^4 - b(1)^3 + c(1)^2 + d(1) + e = 0 \][/tex]
Simplifying, we get:
[tex]\[ a - b + c + d + e = 0 \quad \text{(Equation 1)} \][/tex]
2. Substitute [tex]\( x = -1 \)[/tex] into the polynomial:
[tex]\[ a(-1)^4 - b(-1)^3 + c(-1)^2 + d(-1) + e = 0 \][/tex]
Simplifying, we get:
[tex]\[ a + b + c - d + e = 0 \quad \text{(Equation 2)} \][/tex]
3. Add Equation 1 and Equation 2:
[tex]\[ (a - b + c + d + e) + (a + b + c - d + e) = 0 + 0 \][/tex]
[tex]\[ 2a + 2c + 2e = 0 \][/tex]
Simplifying, we get:
[tex]\[ a + c + e = 0 \][/tex]
4. Subtract Equation 2 from Equation 1:
[tex]\[ (a - b + c + d + e) - (a + b + c - d + e) = 0 - 0 \][/tex]
[tex]\[ -2b + 2d = 0 \][/tex]
Simplifying, we get:
[tex]\[ -b + d = 0 \][/tex]
Thus, we have proven that [tex]\( a + c + e = 0 \)[/tex] and [tex]\( -b + d = 0 \)[/tex].
### Question 16:
To find the value of [tex]\( b \)[/tex] for which [tex]\( (2x + 3) \)[/tex] is a factor of [tex]\( 2x^3 + 9x^2 - x - b \)[/tex]:
1. Factor Theorem: If [tex]\( (2x + 3) \)[/tex] is a factor, then [tex]\( 2x + 3 = 0 \implies x = -\frac{3}{2} \)[/tex] will satisfy the polynomial.
2. Substitute [tex]\( x = -\frac{3}{2} \)[/tex] into the polynomial:
[tex]\[ 2\left(-\frac{3}{2}\right)^3 + 9\left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right) - b = 0 \][/tex]
3. Calculate each term:
[tex]\[ 2\left(-\frac{27}{8}\right) + 9\left(\frac{9}{4}\right) + \frac{3}{2} - b = 0 \][/tex]
[tex]\[ -\frac{54}{8} + \frac{81}{4} + \frac{3}{2} - b = 0 \][/tex]
[tex]\[ -\frac{27}{4} + \frac{81}{4} + \frac{3}{2} - b = 0 \][/tex]
[tex]\[ \frac{54}{4} + \frac{3}{2} - b = 0 \][/tex]
[tex]\[ \frac{27}{2} + \frac{3}{2} - b = 0 \][/tex]
[tex]\[ \frac{30}{2} - b = 0 \][/tex]
[tex]\[ 15 - b = 0 \][/tex]
[tex]\[ b = 15 \][/tex]
Thus, the value of [tex]\( b \)[/tex] for which [tex]\( 2x + 3 \)[/tex] is a factor of [tex]\( 2x^3 + 9x^2 - x - b \)[/tex] is [tex]\( b = 15 \)[/tex].
