Question

Find the zero of the
polynomial
3 x³ - 4 x² + 8x - 7​

Answer 1

Answer:

if x = 1 be the zero of the polynomial, then when we substitute this value in the polynomial, then, the remainder would be zero.

\begin{gathered}x = 1 \\ p(x) = 3 {x}^{3} - 4 {x}^{2} + 8x - 7 \\ p(1) = 3 {(1)}^{3} - 4 {(1)}^{2} + 8(1) - 7 \\ p(x) = (3 \times 1) - (4 \times 1) + 8 - 7 \\ p(x) = 3 - 4 + 8 - 7 \\ p(x) = - 1 + 1 \\ p(x) = 0\end{gathered}

x=1

p(x)=3x

3

−4x

2

+8x−7

p(1)=3(1)

3

−4(1)

2

+8(1)−7

p(x)=(3×1)−(4×1)+8−7

p(x)=3−4+8−7

p(x)=−1+1

p(x)=0

Hence, x = 1 is a zero of polynomial p(x) 3x³-4x²+8x-7

Answer 2

Required Answer:-

Given:

• 3x³ - 4x² + 8x - 7 = 0

To find:

• The zeros of the given polynomial.

Solution:

We have,

➡ 3x³ - 4x² + 8x - 7 = 0

➡ 3x³ - 3x² - x² + x + 7x - 7 = 0

➡ 3x²(x - 1) - x(x - 1) + 7(x - 1) = 0

➡ (x - 1)(3x² - x + 7) = 0

By zero product rule,

★ Either (x - 1) = 0 or (3x² - x + 7) = 0

➡ x = 1

Hence, one of the roots of the given equation is. 1

Now,

➡ 3x² - x + 7 = 0

★ Here,

a = 3

b = -1

c = 7

So,

x = (-b ± √(b² - 4ac))/2a

= (1 ± √(1 - 84))/6

= (1 ± i√83)/6

Hence, the roots of the given Quadratic Equation are (1 + i√83)/6 and (1 - i√83)/6

★ So, the roots of the given cubic equation are 1, (1 + i√83)/6 and (1 - i√83)/6

Answer:

• The roots of the given cubic equation are 1, (1 + i√83)/6 and (1 - i√83)/6

Note:

• i = √-1 (imaginary number)