Question

What is the value of P and Q ​

Answer 1

Answer:

The value of p and q are 1 and -8 receptively.

Step-by-step explanation:

The equation is: $px^{4}+2x^{3}-3x^{2}+qx-4=0$.

One of the factors of this equation is $(x^{2}-4)$.

If $(x^{2}-4)=0$ the value of x are:

$(x^{2}-4)=0\\x=\sqrt{4}\\x=\pm2$

So the value of x is either 2 or -2.

• Substitute x = 2 in the provided equation and solve:

        $px^{4}+2x^{3}-3x^{2}+qx-4=0\\(p\times(2)^{4})+(2\times(2)^{3})-(3\times(2)^{2})+(q\times2)-4=0\\16p+16-12+2q-4=0\\16p+2q=0...(i)$

• Substitute x = -2 in the provided equation and solve:

        $px^{4}+2x^{3}-3x^{2}+qx-4=0\\(p\times(-2)^{4})+(2\times(-2)^{3})-(3\times(-2)^{2})+(q\times-2)-4=0\\16p-16-12-2q-4=0\\16p-2q=32...(ii)$

Add equations (i) and (ii) to eliminate q as follows:

$32p=32\\p=1$

Substitute p = 1 in equation (i) and solve for q as follows:

$16p+2q=0\\(16\times1)+2q=0\\2q=-16\\q=-8$

Thus, the value of p and q are 1 and -8 receptively.