Question:
Let \(\alpha,\beta,\gamma,\delta\) be the roots of
\[
4x^4+8x^3-17x^2-12x+9=0.
\]
If
\[
4(\alpha+4)(\beta+4)(\gamma+4)(\delta+4)=k,
\]
then \(k=\)
Let \(\alpha,\beta,\gamma,\delta\) be the roots of
\[
4x^4+8x^3-17x^2-12x+9=0.
\]
If
\[
4(\alpha+4)(\beta+4)(\gamma+4)(\delta+4)=k,
\]
then \(k=\)
Show Hint
For expressions involving products of \((\alpha+c)\), substitute \(x=-c\) directly into the polynomial.
Updated On: Jun 18, 2026
- \(25\)
- \(35\)
- \(297\)
- \(105\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
For a polynomial
\[
P(x)=a(x-\alpha)(x-\beta)(x-\gamma)(x-\delta),
\]
\[
(\alpha+4)(\beta+4)(\gamma+4)(\delta+4)
=
\frac{P(-4)}{a}.
\]
Step 1: Evaluate \(P(-4)\).
\[ P(x)=4x^4+8x^3-17x^2-12x+9 \] \[ P(-4) = 4(256)+8(-64)-17(16)+48+9 \] \[ =1024-512-272+48+9 \] \[ =297 \]
Step 2: Use the root-product formula.
Leading coefficient \[ a=4 \] Hence \[ (\alpha+4)(\beta+4)(\gamma+4)(\delta+4) = \frac{297}{4} \] Therefore \[ k = 4\times\frac{297}{4} = 297 \] \[ \boxed{297} \]
Step 1: Evaluate \(P(-4)\).
\[ P(x)=4x^4+8x^3-17x^2-12x+9 \] \[ P(-4) = 4(256)+8(-64)-17(16)+48+9 \] \[ =1024-512-272+48+9 \] \[ =297 \]
Step 2: Use the root-product formula.
Leading coefficient \[ a=4 \] Hence \[ (\alpha+4)(\beta+4)(\gamma+4)(\delta+4) = \frac{297}{4} \] Therefore \[ k = 4\times\frac{297}{4} = 297 \] \[ \boxed{297} \]
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