Question

plz answer 15th and 26th its urgent plzzz

Answer 1

$\textbf{ \large{ \underline{ Hello Mate! }}}$

15. ( Error : It will be a + b = 10 )

Solution :

${a}^{2} + {b}^{2} = 58 \\ a + b = 10$

On squaring both sides we get :

${(a + b)}^{2} = {10}^{2} \\ {a}^{2} + {b}^{2} + 2ab = 100 \\ 2ab = 100 - 58 \\ ab = \frac{42}{2} = 21 \\ {a}^{3} + {b}^{3} = (a + b)( {a}^{2} + {b}^{2} - ab) \\ = (10)(58 - 21) \\ = 10 \times 37 = > 370$

$\boxed{ a^3 + b^3 = 370 }$

16. If ( x - 1 ) is factor of p(x) then x = 1

$p(x) = 3 {x}^{3} - {x}^{2} - 3x + 1 \\ p(1) = 3( {1}^{3} ) - {(1)}^{2} - 3(1) + 1 \\ = 3 - 1 - 3 + 1 \\ = 0$

Hence, remainder is "0" therefore ( x - 1 ) is factor if p(x).

$\texttt{ Have great future ahead! }$