Question

If the two zeros of the polynomial(x^4+2x^3-17x^2-18x+72)are 2 and -4, find the value of the remaining zeros.​

Answer 1

Given ;

• p(x) = x⁴ + 2x³ -17x² -18x + 72
• α = 2 (root)
• β = -4 (root)

Thus factors would be :

• (x-2) & (x+4)

Another factor would be :

$(x - 2) \times (x + 4 )\\ {x}^{2} + 2x - 8$

By long division method when we divide p(x) by x² + 2x - 8

We get quotient =

• x² - 9
• (x² -9)
• (x-3)(x+3)

Thus the value of remaining zeros ;

when x -3 = 0

• x = 3 = γ (other root)

when x + 3 = 0

• x = -3 = δ (other root)

Thus all the roots of this p(x) would be ;

$\alpha = 2 \\ \beta = - 4 \\ \gamma = 3 \\ \delta = - 3$

#answerwithquality

#BAL