(i) $2x^3 + x^2 – 5x + 2;$ $\dfrac{1}{2}, 1, – 2$
(ii) $x^3-4x^2+5x-2$; 2, 1, 1
Q1. (i) $2x^3 + x^2 – 5x + 2;$ $\dfrac{1}{2}, 1, – 2$
Given,
p(x) = $2x^3+x^2-5x+2$
zeroes for p(x) are =$\dfrac{1}{2}$, 1, -2
$p(\dfrac{1}{2})= 2(\dfrac{1}{2})^{3} + (\dfrac{1}{2})^{2}-5(\dfrac{1}{2})+2 = \dfrac{1}{4}+\dfrac{1}{4}-\dfrac{5}{2}+2 = 0$
$p(1) = 2(1)^3+(1)^2-5(1)+2 = 0$
$p(-2) = 2(-2)^3+(-2)^2-5(-2)+2 = 0$
Hence, Prooved $\dfrac{1}{2}$, 1, -2 are the zeroes of $2x^3+x^2-5x+2$.
Comparing the given polynomial with $ax^3+bx^2+cx+d$, we get
a = 2, b = 1, c = – 5, d = 2
α = $\dfrac{1}{2}$
β = 1
γ = -2
As we know, if α, β, γ are the zeroes of the cubic polynomial $ax^3+bx^2+cx+d$ , then;
α+β+γ = $\dfrac{–b}{a}$
$\dfrac{1}{2}+ 1+(-2) = \dfrac{-1}{2}$
$\dfrac{-1}{2} = \dfrac{-1}{2}$
αβ+βγ+γα = $\dfrac{c}{a}$
=>$(\dfrac{1}{2} \times 1) + (1 \times -2)+(-2 \times \dfrac{1}{2}) = \dfrac{–5}{2}$
$\dfrac{–5}{2} = \dfrac{–5}{2}$
αβγ = $\dfrac{–d}{a}$
=>αβγ = $\dfrac{1}{2} \times 1 \times (-2) = \dfrac{-2}{2} $
$\dfrac{-2}{2} = \dfrac{-2}{2} $
Hence, the relationship between the zeroes and the coefficients is Verified
(ii) $x^3-4x^2+5x-2$; 2, 1, 1
Given, p(x) = $x^3-4x^2+5x-2$
zeroes for p(x) are 2,1,1
p(2)= $2^3-4(2)^2+5(2)-2 = 0$
p(1) = $1^3-(4×1^2 )+(5×1)-2 = 0$
Hence proved, 2, 1, 1 are the zeroes of $x^3-4x^2+5x-2$
a=1, b=−4,c=5, d=−2
α=2
β=1
γ=1
As we know, if α, β, γ are the zeroes of the cubic polynomial $ax^3+bx^2+cx+d$ , then;
α + β + γ = $\dfrac{–b}{a}$
2 + 1 + 1 = $\dfrac{-(–4)}{1}$
4 = 4
αβ+βγ+γα = $\dfrac{c}{a}$
(2×1) + (1×1) + (1×2) = $\dfrac{5}{1}$
5 = $\dfrac{5}{1}$
αβγ = $\dfrac{–d}{a}$
2×1×1 = $\dfrac{–(-2)}{1}$
2 = $\dfrac{2}{1}$
Hence, the relationship between the zeroes and the coefficients is Verified
Let us consider the cubic polynomial is $ax^3+bx^2+cx+d$, and the values of the zeroes of the polynomials are α, β, γ.
As per the given question,
$ x^3+2x^2-7x-14$
a = 1
b = 2
c = -7
d = -14
α + β + γ = $\dfrac{–b}{a} = \dfrac{–2}{1}$
αβ+βγ+γα = $\dfrac{c}{a} = \dfrac{-7}{1}$
αβγ = = $\dfrac{–d}{a} = \dfrac{–(-14)}{1}$
Thus, from the above three expressions, we get the values of the coefficients of the polynomial.
a = 1, b = -2, c = -7, d = 14
Hence, the cubic polynomial is $x^3-2x^2-7x+14$
p(x) = $x^3-3x^2+x+1$
zeroes are given as a – b, a, a + b
comparing the given polynomial with the general expression
$px^3+qx^2+rx+s = x^3-3x^2+x+1$
p = 1, q = -3, r = 1 and s = 1
Sum of zeroes = $\dfrac{-(Coefficient\:\:of\:\:x^{2})}{(Coefficient \:\:of \:\:x^{3})}= \dfrac{-q}{p}$
a – b + a + a + b = $\dfrac{-3}{1}$
3a = $\dfrac{-(-3)}{1}$
a = 1
Thus, the zeroes are 1-b, 1, 1+b.
Product of zeroes = $\dfrac{Constant \:term}{(Coefficient \:\: of \:\: x^{3})} = = \dfrac{-s}{p}$
$ (1-b) \times 1 \times (1+b) = \dfrac{-1}{1}$
$ 1 - b^{2} = -1$
$ b^{2} $= 1 + 1 = 2
$b = ± \sqrt{2}$
Hence,1-$\sqrt{2}$, 1 ,1+$\sqrt{2}$ are the zeroes of $x^3-3x^2+x+1$.
Since this is a polynomial equation of degree 4, there will be total of 4 roots
Let f(x) = $x^4-6x^3-26x^2+138x-35$
Since $2 + \sqrt{3}$ and $2-\sqrt{3}$ are zeroes of given polynomial f(x)
$∴ [x−(2+\sqrt{3})] [x−(2-√3)] = 0$
$(x−2−\sqrt{3})(x−2+\sqrt{3}) = 0$
$x^2-2x+\sqrt{3}x-2x+4-2\sqrt{3}-\sqrt{3}x+2\sqrt{3}-3$
$x^2-4x+1$ , this is a factor of a given polynomial f(x)
Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x), and the remainder will be 0
So, $x^4-6x^3-26x^2+138x-35 = (x^2-4x+1)(x^2 –2x−35)$
Now, on further factorizing (x2–2x−35) we get,
$x^2–(7−5)x −35 = x^2– 7x+5x+35 = 0$
x(x −7)+5(x−7) = 0
(x+5)(x−7) = 0
So, its zeroes are given by:
x= −5 and x = 7
Therefore, all four zeroes of the given polynomial equation are $2+\sqrt{3}$ , $2-\sqrt{3}$, −5 and 7
Let’s divide $x^4 – 6x^3 + 16x^2 – 25x + 10$ by $x^2 – 2x + k$
Given that the remainder of the polynomial division is x + a
(4k – 25 + 16 – 2k)x + [10 – k(8 – k)] = x + a
$(2k – 9)x + (10 – 8k + k^2) = x + a$
Comparing the coefficients of the above equation, we get;
2k – 9 = 1
2k = 9 + 1 = 10
k = $\dfrac{10}{2}$ = 5
And
$10 – 8k + k^2 = a$
$10 – 8(5) + (5)^2 = a [since k = 5]$
10 – 40 + 25 = a
a = -5
Therefore, k = 5 and a = -5