On diminishing the roots of $x^{5} + 4 x^{3} - x^{2} + 11 = 0$ by $3$ , the transformed equation is $y^{5} + p_{1} y^{4} + p_{2} y^{3} + p_{3} y^{2} + p_{4} y + p_{5} = 0$ , then $p_{3} =$

353

No worries! We‘ve got your back. Try BYJU‘S free classes today!

507

No worries! We‘ve got your back. Try BYJU‘S free classes today!

305

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

94

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $305$
$f (x) = x^{5} + 4 x^{3} - x^{2} + 11$
The solutions of the new equation are $3$ less than that of $f (x) = 0$ .
So the new equation is $f (y + 3) = 0$ .
$f (y + 3) = (y + 3)^{5} + 4 (y + 3)^{3} - (y + 3)^{2} + 11 = 0$
We need to find the coefficient of $y^{2}$ in the expansion of $f (y + 3) = 0$ .
So we take the sum of coefficients of $y^{2}$ from each of the above terms.
$p_{3} =^{5} C_{2} \times 3^{3} + 4 \times^{3} C_{2} \times 3 - 1 = 305.$

Suggest Corrections

Similar questions

x^{n} + p_{1} x^{n - 1} + p_{2} x^{n - 2} . . . . . + P_{n - 1} x + P_{n} = 0 (p_{1}, p_{2}, p_{3} . . . . a r e r e a l), t h e n p r o v e t h a t (1 + a^{2}) (1 + b^{2}) . . . (1 + k^{2}) = (1 - p_{2} + p_{4} - . . . .)^{2} + (p_{1} - p_{3} + p_{5} - . .)^{2}

{(1 + x)}^{n} = P_{0} + P_{1} x + P_{2} x^{2} + . . . + P_{n} x^{n}

\frac{p_{0} - p_{2} + p_{4} - p_{6} + . . .}{p_{1} - p_{3} + p_{5} - p_{7} + . . .} = ?

P (x)

4

P (1) = 2, P (2) = 5, P (3) = 10

P (4) = 17

P (5) =

a, b, c, . . . k

x^{n} + p_{1} x^{n - 1} + p_{2} x^{n - 2} + . . . . + p_{n - 1} x + p_{n} = 0

(1 + a^{2}) (1 + b^{2}) . . . . (1 + k^{2}) = (1 - p_{2} + p_{4} - . . . .)^{2} + (p_{1} - p_{3} + p_{5} - . . . .)^{2}

P_{1} = P_{2}

P_{2} = P_{3}

P_{3} = P_{4}

P_{1} = P_{4}

Join BYJU'S Learning Program