# Answer to Question #10704 in Algebra for Manthan khorwal

Question #10704

Find the value of a and b so that (x+2) and (x-2) are the factors of p(x)=ax4+2x3-3x2+bx-4.

Expert's answer

(x + 2) * (x - 2) = x^2 - 4

ax^4 + 2x^3 - 3x^2 + bx - 4 | x^2 -

4

ax^4 - 4ax^2 | ax^2 + 2x + (4a-3)

2x^3 + (-3+4a)x^2

2x^3 - 8x

(4a-3)x^2

+ (b+8)x

(4a-3)x^2 +

(-16a+12)

(b+8)x - 4 + 16a - 12 =

0

{ b + 8 = 0, -4 + 16a -12 = 0 }, { a = 1, b = -8 }

ax^4 + 2x^3 - 3x^2 + bx - 4 | x^2 -

4

ax^4 - 4ax^2 | ax^2 + 2x + (4a-3)

2x^3 + (-3+4a)x^2

2x^3 - 8x

(4a-3)x^2

+ (b+8)x

(4a-3)x^2 +

(-16a+12)

(b+8)x - 4 + 16a - 12 =

0

{ b + 8 = 0, -4 + 16a -12 = 0 }, { a = 1, b = -8 }

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