Question

$p(x) = - 3x{3} + kx2 - 8x + 10 \\ zero = 2 \\ find \: k$
​

Answer 1

Given,

$\rm{}p(x) = { - 3x}^{3} + {kx}^{2} - 8x + 10$

• One of Zeroes of this polynomial is 2

To find : The value of k

↓ Solution →

Putting the value of x in p(x), we get,

$\rm \to { - 3(2)}^{3} + {k(2)}^{2} - 8(2) + 10 = 0 \\ \rm \to \: \: - 24 + 4k - 16 + 10 = 0 \\ \rm \to \: \: 4k + (10 - 24 - 16) = 0 \\ \rm \to \: \: \: \: 4k = \: \: - (10 - 24 - 16) \\ \rm \to \: \: \: \: \: 4k = \: \: \: - (10 - 40) \\ \rm \to \: \: \: \: 4k = \: \: 40 - 10 \\ \rm \to \: \: \: \: \: 4k = \: \: 30 \\ \\ \rm \to \frac{4k}{4} = \frac{30}{4} \\ \\ \rm \to \: \: \: \: k = 7\frac{2}{4} \\ \\ \rm \to \: \: \: \: \: k = 7 \frac{1}{2} \\ \\ \rm \to \boxed{ \sf\: \: \: \: \: k = 7.5}$

Answer 2

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Given,

$\rm{}p(x) = { - 3x}^{3} + {kx}^{2} - 8x + 10$

One of Zeroes of this polynomial is 2

To find : The value of k

↓ Solution →

Putting the value of x in p(x), we get,

$\rm \to { - 3(2)}^{3} + {k(2)}^{2} - 8(2) + 10 = 0 \\ \rm \to \: \: - 24 + 4k - 16 + 10 = 0 \\ \rm \to \: \: 4k + (10 - 24 - 16) = 0 \\ \rm \to \: \: \: \: 4k = \: \: - (10 - 24 - 16) \\ \rm \to \: \: \: \: \: 4k = \: \: \: - (10 - 40) \\ \rm \to \: \: \: \: 4k = \: \: 40 - 10 \\ \rm \to \: \: \: \: \: 4k = \: \: 30 \\ \\ \rm \to \frac{4k}{4} = \frac{30}{4} \\ \\ \rm \to \: \: \: \: k = 7\frac{2}{4} \\ \\ \rm \to \: \: \: \: \: k = 7 \frac{1}{2} \\ \\ \rm \to \boxed{ \sf\: \: \: \: \: k = 7.5}$