1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x³ + x² – 5x + 2; ¹/₂ , 1, -2

(ii) x³ – 4x² + 5x – 2; 2,1,1

Solution:

(i) 2x³ + x² – 5x + 2; ¹/₂ , 1, -2

p(x) = 2x³ + x² – 5x + 2

Zeros for this polynomial are ¹/₂ , 1, -2

p(¹/₂) = 2(¹/₂)³ + (¹/₂)² – 5(¹/₂) + 2

= (¹/₄) + (¹/₄) – (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

= -16 + 4 + 10 + 2

= 0

Therefore, ½ , 1, and −2 are the zeroes of the given polynomial. Comparing the given polynomial with ax³ + bx² + cx + d, we obtain a = 2, b = 1, c = −5, d = 2

Let us take α = ¹/₂ ,β = 1, γ = -2

α + β + γ = ¹/₂ + 1 + (-2) = –¹/₂ = –^b/_a

αβ + βγ + γα = ¹/₂ x 1 + 1(-2) + ¹/₂ (-2) = ^-5/₂ = ^c/_a

αβγ  = ¹/₂ x 1 x (-2) = –¹/₁ = – ²/₂ = –^d/_a

Therefore, the relationship between the zeroes and the coefficients is verified.

 

(ii) x³ – 4x² + 5x – 2; 2,1,1

p(x) = x³ – 4x² + 5x – 2

Zeros for this polynomial are 2 , 1, 1

p(2) = 2³ – 4(2)² + 5(2) – 2

= 8 – 16 + (10) – 2

= 0

p(1) = (1)³ – 4(1)² + 5(1) – 2 = 0

Therefore, 2 , 1, and 1 are the zeroes of the given polynomial. Comparing the given polynomial with ax³ + bx² + cx + d, we obtain a = 1, b = -4, c = 5, d = -2

Verify:

Sum of zeros = 2+1+1 = 4 = – ^(-4)/₁ = –^b/_a

Multiplication of zeroes taking two at a time = (2)(1)+(1)(1)+(2)(1) = 2+1+2= 5 = ⁵/₁ = ^c/_a

Multiplication of zeroses = 2 x 1 x 1 = 2 = – ^(-2)/₁ = –^d/_a

Hence, the relationship between the zeroes and the coefficients is verified.

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2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, − 7, − 14 respectively. 

Solution:

Let the polynomial be ax³ + bx² + cx + d and the zeroes be α, β and γ.

It is given that

α + β + γ = ²/₁ = –^b/_a

αβ + βγ + γα = ^-7/₂ = ^c/_a

αβγ  = – ¹⁴/₁ = –^d/_a

If a = 1, then b = −2, c = −7, d = 14 Hence, the polynomial is x³ – 2x² – 7x + 14

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3. If the zeroes of polynomial, x³ -3x²+x+1 are a-b,a,a+b. Find a and b.

Solution:

p(x) = x³ -3x²+x+1

Zeroes are a − b, a + a + b Comparing the given polynomial with px³+qx²+rx+1, we obtain p = 1, q = −3, r = 1, t = 1

Sum of zeros = a – b + a + a + b

^-q/ _p = 3a

–^(-3)/₁  = 3a

3 = 3a

a = 1

The zeroes are 1-b, 1, 1+b

Multiplication of zeroes = 1(1-b)(1+b)

^-t/_p = 1 – b²

^-1/₁= 1 – b²

1 – b² = -1

1 + 1 = b²

b = ±√2

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4. If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ±√3 , find other zeroes.

Solution:

Given 2 + √3 and 2 – √3 are zeroes of the given polynomial. So, (2 + √3)(2 – √3) is a factor of polynomial.

Therefore,[x – (2 + √3)][x – (2 – √3)] = x² + 4 − 4x − 3 = x² − 4x + 1 is a factor of the given polynomial

For finding the remaining zeroes of the given polynomial, we will find

Clearly, x⁴ -6x³ -26x²+138x -35 = (x² – 4x + 1)(x² – 2x – 35)

We know, (x² – 2x – 35) = (x – 7)(x + 5)

Therefore, the value of the polynomial is also zero when (x -7  = 0) or (x + 5 = 0)

Or x = 7 or −5

therefore, 7 and −5 are also zeroes of this polynomial.

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5. If the polynomial x⁴  – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.

Solution:

By division algorithm we know,

Dividend = Divisor × Quotient + Remainder

Dividend − Remainder = Divisor × Quotient

x⁴ – 6x³ + 16x² – 25x + 10 – x  – a = x⁴ – 6x³ + 16x² – 26x  + 10 – a will be divisible by x²  – 2x + k

It ccan be observed that (-10+2k)x +(10-a-8k+k²) will be zero.

Thus, (-10 + 2k) = 0 and (10 – a – 8k +k²) = 0

For (-10+2k) = 0 , 2k = 10 and thus k = 5

For (10 – a – 8k + k²) = 0

10 − a − 8 × 5 + 25 = 0

10 − a − 40 + 25 = 0

− 5 − a = 0

Therefore, a = −5

Hence, k = 5 and a = −5

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