Question

Find the value of k if t-1 is the factor of p(t)=4t^3+3t^2-3t+k

Answer 1

$\large \bf \green {✧~Given ~✧}$

$\sf t-1$ is the factor of $\sf 4t^3 +3t^2 -3t +k$.

$\large \bf \green {✧~To~find~✧}$

We have to find the value of k when t-1 is the factor of the polynomial.

$\large \bf \green {✧~Solution~✧}$

t-1 is the factor of the polynomial, so substituting t=1 in the equation and solving :-

$\pink \implies \sf {4t}^{3} + {3t}^{2} - 3t + k = 0$

$\pink \implies \sf (4) {(1)}^{3} + (3) {(1)}^{2} - 3(1) + k = 0$

$\pink \implies \sf \: 4(1) + 3(1) - 3 + k = 0$

$\pink \implies \sf \: 4 + 3 - 3 +k = 0$

$\pink \implies \sf 4 + k = 0$

$\pink \implies \sf \: k = - 4$

▪️ Therefore, k = -4 for the polynomial ${\bold {\sf 4t^3 +3t^2 -3t +k}}$ when t-1 is a factor of it.

$\large \bf \green {✧~Verification~✧}$

Substituting t-1 and k = -4 in the equation and solving :-

$\pink \implies \sf {4t}^{3} + {3t}^{2} - 3t + k = 0$

$\pink \implies \sf4 {(1)}^{3} + 3 {(1)}^{2} - 3(1) + ( - 4) = 0$

$\pink \implies \sf4 + 3 - 3 - 4 = 0$

$\pink \implies \sf 0 = 0$

▪️LHS = RHS

▪️ Hence, verified !!

Answer 2

Step-by-step explanation:

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