Question

Find the numbers a, b and c if the polynomial x3−18x2+24xb−c is the cube of the binomial x+2a.

Answer 1

Answer:

$a=-3\\ \\b=\dfrac{9}{2}\\ \\c=216$

Step-by-step explanation:

Use formula

$(u+v)^3=u^3+3u^2v+3uv^2+v^3.$

Hence,

$(x+2a)^3=x^3+3x^2\cdot 2a+3x\cdot (2a)^2+(2a)^3=x^3+6ax^2+12a^2x+8a^3.$

This expression is equal to

$x^3-18x^2+24bx-c,$

so we can equate the coefficients at powers of x:

$x^3:\ 1=1\\ \\x^2:\ 6a=-18\Rightarrow a=-3\\ \\x:\ 12a^2=24b\Rightarrow b=\dfrac{9}{2}\\ \\1=x^0:\ 8a^3=-c\Rightarrow c=-8\cdot (-3)^3=216.$