Answer :
Sure, let’s go through each question step-by-step:
### Q4. Find the integral zeros of [tex]\(4x^3 + 20x^2 - x - 5\)[/tex].
To find the integral zeros (roots) of the polynomial [tex]\(4x^3 + 20x^2 - x - 5\)[/tex], we use the Rational Root Theorem. This theorem states that any potential rational root, expressed in simplified form [tex]\(\frac{p}{q}\)[/tex], where [tex]\(p\)[/tex] is a factor of the constant term [tex]\(-5\)[/tex] and [tex]\(q\)[/tex] is a factor of the leading coefficient [tex]\(4\)[/tex].
Factors of [tex]\(-5\)[/tex]: [tex]\(\pm 1, \pm 5\)[/tex]
Factors of [tex]\(4\)[/tex]: [tex]\(\pm 1, \pm 2, \pm 4\)[/tex]
Possible rational roots: [tex]\(\frac{\pm 1}{\pm 1}, \frac{\pm 5}{\pm 1}, \frac{\pm 1}{\pm 2}, \frac{\pm 5}{\pm 2}, \frac{\pm 1}{\pm 4}, \frac{\pm 5}{\pm 4}\)[/tex]
These simplify to the potential candidates: [tex]\(\pm 1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{4}, \pm \frac{5}{4}\)[/tex].
Testing these values, we find:
For [tex]\(x = 1\)[/tex],
[tex]\[4(1)^3 + 20(1)^2 - 1 - 5 = 4 + 20 - 1 - 5 = 18\][/tex]
For [tex]\(x = -1\)[/tex],
[tex]\[4(-1)^3 + 20(-1)^2 - (-1) - 5 = -4 + 20 + 1 - 5 = 12\][/tex]
Neither [tex]\(1\)[/tex] nor [tex]\(-1\)[/tex] are zeros. Upon further testing, we find that [tex]\(x = -\frac{1}{2}\)[/tex] satisfies the equation:
[tex]\[4\left(-\frac{1}{2}\right)^3 + 20\left(-\frac{1}{2}\right)^2 -\left(-\frac{1}{2}\right) -5 = -\frac{4}{8} + \frac{20}{4} + \frac{1}{2} - 5 = -\frac{1}{2} + 5 + \frac{1}{2} - 5 =0.\][/tex]
Therefore, the integral zero is [tex]\(\boxed{-\frac{1}{2}}\)[/tex].
### Q5. Show that the polynomial [tex]\(x^2 + 2x + 7\)[/tex] has no zeros.
To determine if [tex]\(x^2 + 2x + 7\)[/tex] has zeros, we calculate its discriminant. The discriminant ([tex]\(\Delta\)[/tex]) for a quadratic equation [tex]\(ax^2 + bx + c\)[/tex] is given by:
[tex]\[\Delta = b^2 - 4ac\][/tex]
For the given polynomial [tex]\(x^2 + 2x + 7\)[/tex], we have [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = 7\)[/tex].
[tex]\[\Delta = 2^2 - 4 \cdot 1 \cdot 7 = 4 - 28 = -24\][/tex]
Since the discriminant is less than zero ([tex]\(-24 < 0\)[/tex]), the quadratic equation has no real zeros. Hence, the polynomial [tex]\(x^2 + 2x + 7\)[/tex] has no zeros.
### Q6. Without actual division, prove that [tex]\(2x^4 - 5x^3 + 2x^2 - x + 2\)[/tex] is divisible by [tex]\(x^2 - 3x + 2\)[/tex].
First, factorize [tex]\(x^2 - 3x + 2\)[/tex]:
[tex]\[x^2 - 3x + 2 = (x-1)(x-2)\][/tex]
Now substitute [tex]\(x = 1\)[/tex] and [tex]\(x = 2\)[/tex] into [tex]\(2x^4 - 5x^3 + 2x^2 - x + 2\)[/tex] to show both satisfy the polynomial:
For [tex]\(x = 1\)[/tex]:
[tex]\[2(1)^4 - 5(1)^3 + 2(1)^2 - 1 + 2 = 2 - 5 + 2 - 1 + 2 = 0\][/tex]
For [tex]\(x = 2\)[/tex]:
[tex]\[2(2)^4 - 5(2)^3 + 2(2)^2 - 2 + 2 = 32 - 40 + 8 - 2 + 2 = 0\][/tex]
Since both values satisfy the polynomial, [tex]\(2x^4 - 5x^3 + 2x^2 - x + 2\)[/tex] is divisible by [tex]\(x^2 - 3x + 2\)[/tex].
### Q7. If the polynomial [tex]\(2x^3 - ax^2 + bx + 4\)[/tex] has [tex]\(x+1\)[/tex] as a factor and leaves remainder 4 when divided by [tex]\(2x+1\)[/tex], find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
Since [tex]\(x+1\)[/tex] is a factor, [tex]\(P(-1) = 0\)[/tex]:
[tex]\[2(-1)^3 - a(-1)^2 + b(-1) + 4 = 0\Rightarrow -2 - a - b + 4 = 0 \Rightarrow -a - b + 2 = 0 \Rightarrow a + b = 2 \][/tex]
Since the remainder when divided by [tex]\(2x+1\)[/tex] is 4, substitute [tex]\(x = -\frac{1}{2}\)[/tex]:
[tex]\[2\left(-\frac{1}{2}\right)^3 - a\left(-\frac{1}{2}\right)^2 + b\left(-\frac{1}{2}\right) + 4 = 4 \Rightarrow -\frac{2}{8} -\frac{a}{4} - \frac{b}{2} + 4 = 4 \Rightarrow -\frac{1}{4} - \frac{a}{4} - \frac{b}{2} + 4 = 4\Rightarrow - a-2b +1= 0\Rightarrow 4a=7b-4\Rightarrow 8a= 14b-8\][/tex]
Solving the system:
[tex]\[a + b = 2\\ 4a + 2b = 8.\][/tex]
Thus, the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are [tex]\(a=0\)[/tex] and [tex]\(b=2\ ). ### Q8. Factorise the polynomial \(x^4 + 2x^3 - 13x^2 - 14x + 24\)[/tex].
We use the Rational Root Theorem to find possible rational roots: [tex]\(\pm 1, \pm 2,\pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\)[/tex].
Testing [tex]\(x = 2\)[/tex]:
[tex]\[2^4 + 2 \cdot 2^3 - 13 \cdot 2^2 - 14 \cdot 2 + 24 = 16 + 16 - 52 - 28 + 24 = 0\][/tex]
Thus, [tex]\(x = 2\)[/tex] is a root. Now perform synthetic division or polynomial division to factor [tex]\(x-2\)[/tex] out:
After performing polynomial division, the quotient is:
[tex]\[x^3 + 4x^2 - 5x - 12\][/tex]
Factorize the quotient to find remaining factors:
Testing [tex]\(x = -2\)[/tex]:
[tex]\[(-2)^3 +4(-2)^2 + 5(-2)-12=0\][/tex]
(Use synthetic division again - this factors into polynomial and we get [tex]\((x-2)(x^2-1)\]) \[x^2-(-xy\] Therefore the polynomial is \(x^4 + 2x^3 - 13 x^2 - 14 x + 24 =(x-2)(x \times x+)\)[/tex]
Thus, the fully factorised form of the polynomers \[(x^3)(x-12)
### Q4. Find the integral zeros of [tex]\(4x^3 + 20x^2 - x - 5\)[/tex].
To find the integral zeros (roots) of the polynomial [tex]\(4x^3 + 20x^2 - x - 5\)[/tex], we use the Rational Root Theorem. This theorem states that any potential rational root, expressed in simplified form [tex]\(\frac{p}{q}\)[/tex], where [tex]\(p\)[/tex] is a factor of the constant term [tex]\(-5\)[/tex] and [tex]\(q\)[/tex] is a factor of the leading coefficient [tex]\(4\)[/tex].
Factors of [tex]\(-5\)[/tex]: [tex]\(\pm 1, \pm 5\)[/tex]
Factors of [tex]\(4\)[/tex]: [tex]\(\pm 1, \pm 2, \pm 4\)[/tex]
Possible rational roots: [tex]\(\frac{\pm 1}{\pm 1}, \frac{\pm 5}{\pm 1}, \frac{\pm 1}{\pm 2}, \frac{\pm 5}{\pm 2}, \frac{\pm 1}{\pm 4}, \frac{\pm 5}{\pm 4}\)[/tex]
These simplify to the potential candidates: [tex]\(\pm 1, \pm 5, \pm \frac{1}{2}, \pm \frac{5}{2}, \pm \frac{1}{4}, \pm \frac{5}{4}\)[/tex].
Testing these values, we find:
For [tex]\(x = 1\)[/tex],
[tex]\[4(1)^3 + 20(1)^2 - 1 - 5 = 4 + 20 - 1 - 5 = 18\][/tex]
For [tex]\(x = -1\)[/tex],
[tex]\[4(-1)^3 + 20(-1)^2 - (-1) - 5 = -4 + 20 + 1 - 5 = 12\][/tex]
Neither [tex]\(1\)[/tex] nor [tex]\(-1\)[/tex] are zeros. Upon further testing, we find that [tex]\(x = -\frac{1}{2}\)[/tex] satisfies the equation:
[tex]\[4\left(-\frac{1}{2}\right)^3 + 20\left(-\frac{1}{2}\right)^2 -\left(-\frac{1}{2}\right) -5 = -\frac{4}{8} + \frac{20}{4} + \frac{1}{2} - 5 = -\frac{1}{2} + 5 + \frac{1}{2} - 5 =0.\][/tex]
Therefore, the integral zero is [tex]\(\boxed{-\frac{1}{2}}\)[/tex].
### Q5. Show that the polynomial [tex]\(x^2 + 2x + 7\)[/tex] has no zeros.
To determine if [tex]\(x^2 + 2x + 7\)[/tex] has zeros, we calculate its discriminant. The discriminant ([tex]\(\Delta\)[/tex]) for a quadratic equation [tex]\(ax^2 + bx + c\)[/tex] is given by:
[tex]\[\Delta = b^2 - 4ac\][/tex]
For the given polynomial [tex]\(x^2 + 2x + 7\)[/tex], we have [tex]\(a = 1\)[/tex], [tex]\(b = 2\)[/tex], and [tex]\(c = 7\)[/tex].
[tex]\[\Delta = 2^2 - 4 \cdot 1 \cdot 7 = 4 - 28 = -24\][/tex]
Since the discriminant is less than zero ([tex]\(-24 < 0\)[/tex]), the quadratic equation has no real zeros. Hence, the polynomial [tex]\(x^2 + 2x + 7\)[/tex] has no zeros.
### Q6. Without actual division, prove that [tex]\(2x^4 - 5x^3 + 2x^2 - x + 2\)[/tex] is divisible by [tex]\(x^2 - 3x + 2\)[/tex].
First, factorize [tex]\(x^2 - 3x + 2\)[/tex]:
[tex]\[x^2 - 3x + 2 = (x-1)(x-2)\][/tex]
Now substitute [tex]\(x = 1\)[/tex] and [tex]\(x = 2\)[/tex] into [tex]\(2x^4 - 5x^3 + 2x^2 - x + 2\)[/tex] to show both satisfy the polynomial:
For [tex]\(x = 1\)[/tex]:
[tex]\[2(1)^4 - 5(1)^3 + 2(1)^2 - 1 + 2 = 2 - 5 + 2 - 1 + 2 = 0\][/tex]
For [tex]\(x = 2\)[/tex]:
[tex]\[2(2)^4 - 5(2)^3 + 2(2)^2 - 2 + 2 = 32 - 40 + 8 - 2 + 2 = 0\][/tex]
Since both values satisfy the polynomial, [tex]\(2x^4 - 5x^3 + 2x^2 - x + 2\)[/tex] is divisible by [tex]\(x^2 - 3x + 2\)[/tex].
### Q7. If the polynomial [tex]\(2x^3 - ax^2 + bx + 4\)[/tex] has [tex]\(x+1\)[/tex] as a factor and leaves remainder 4 when divided by [tex]\(2x+1\)[/tex], find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
Since [tex]\(x+1\)[/tex] is a factor, [tex]\(P(-1) = 0\)[/tex]:
[tex]\[2(-1)^3 - a(-1)^2 + b(-1) + 4 = 0\Rightarrow -2 - a - b + 4 = 0 \Rightarrow -a - b + 2 = 0 \Rightarrow a + b = 2 \][/tex]
Since the remainder when divided by [tex]\(2x+1\)[/tex] is 4, substitute [tex]\(x = -\frac{1}{2}\)[/tex]:
[tex]\[2\left(-\frac{1}{2}\right)^3 - a\left(-\frac{1}{2}\right)^2 + b\left(-\frac{1}{2}\right) + 4 = 4 \Rightarrow -\frac{2}{8} -\frac{a}{4} - \frac{b}{2} + 4 = 4 \Rightarrow -\frac{1}{4} - \frac{a}{4} - \frac{b}{2} + 4 = 4\Rightarrow - a-2b +1= 0\Rightarrow 4a=7b-4\Rightarrow 8a= 14b-8\][/tex]
Solving the system:
[tex]\[a + b = 2\\ 4a + 2b = 8.\][/tex]
Thus, the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are [tex]\(a=0\)[/tex] and [tex]\(b=2\ ). ### Q8. Factorise the polynomial \(x^4 + 2x^3 - 13x^2 - 14x + 24\)[/tex].
We use the Rational Root Theorem to find possible rational roots: [tex]\(\pm 1, \pm 2,\pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24\)[/tex].
Testing [tex]\(x = 2\)[/tex]:
[tex]\[2^4 + 2 \cdot 2^3 - 13 \cdot 2^2 - 14 \cdot 2 + 24 = 16 + 16 - 52 - 28 + 24 = 0\][/tex]
Thus, [tex]\(x = 2\)[/tex] is a root. Now perform synthetic division or polynomial division to factor [tex]\(x-2\)[/tex] out:
After performing polynomial division, the quotient is:
[tex]\[x^3 + 4x^2 - 5x - 12\][/tex]
Factorize the quotient to find remaining factors:
Testing [tex]\(x = -2\)[/tex]:
[tex]\[(-2)^3 +4(-2)^2 + 5(-2)-12=0\][/tex]
(Use synthetic division again - this factors into polynomial and we get [tex]\((x-2)(x^2-1)\]) \[x^2-(-xy\] Therefore the polynomial is \(x^4 + 2x^3 - 13 x^2 - 14 x + 24 =(x-2)(x \times x+)\)[/tex]
Thus, the fully factorised form of the polynomers \[(x^3)(x-12)
