Question

Let P(x) = x² + bx + c, where b and c are integers. If P(x) is a factor of both x⁴ + 6x² + 25 and 3x⁴ + 4x² + 28x + 5, then value of P(1) is -

Answer 1

\( P(x)=x^2+b x+c \text { is a factor of } x^4+6 x^2+25\)

\( x^4+6 x^2+25=x^4+10 x^2-4 x^2+25=\left(x^4+10 x^2+25\right)-4 x^2=\left(x^2+5\right)^2-(2 x)^2 \)

\( \Rightarrow x^4+6 x^2+25=\left(x^2+2 x+5\right)\left(x^2-2 x+5\right)\)

One of them is a factor of \(3 x^4+4 x^2+28 x+5\).

\( 3 x^4+4 x^2+28 x+5=3 x^4-6 x^3+15 x^2+6 x^3-11 x^2+28 x+5 \)

\( \Rightarrow 3 x^2\left(x^2-2 x+5\right)+6 x^3-12 x^2+30 x+x^2-2 x+5=3 x^2\left(x^2-2 x+5\right)+6 x\left(x^2-2 x+5\right)+x^2 \)\( -2 x+5 \)

\( \Rightarrow 3 x^4+4 x^2+28 x+5=\left(x^2-2 x+5\right)\left(3 x^2+6 x+1\right)\)

So, \(P(x)=x^2-2 x+5\)

\(\therefore P(1)=1-2+5=4\)