Factorize the polynomial f=x^3+4x^2-7x-10
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We have to factorize f=x^3 + 4x^2 - 7x - 10
f=x^3 + 4x^2 - 7x - 10
=> x^3 + 6x^2 - 2x^2 + 5x - 12x - 10
=> x^3...
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To factorise the polynomial f(x) = x^3+4x^2-7x-10.
We see that f(2) = 2^3+4*2^2-7*2-10 = 8+16-14-10 = 0.
Therefore we can write f(x) = (x-2)(x^2+kx+5) so that x^3 and constant terms are equal on both sides.
Put x = 1, then f(1) = 1^3+4*1^2-7*1-10 = (1-2)((1^2+k*1+5)
-12 = (-1)(6+k) = k-6. So k = 12-6 = 6.
Therefore x^3+4x^2-7x-10 =(x-2)(x^2+6x+5)....(1)
x^2-6x+5 = (x+5)(x+1).
So f(x) = x^3+4x^2-7x-10 = (x-2)(x+1)(x-5).
Therefore the factors of the polynomial
x^3+4x^2-7x-10 = (x-2)(x+1)(x+5).
First, we'll verify if x = -1 is the root of polynomial.
f(-1) = (-1)^3 + 4*(-1)^2 - 7*(-1) - 10
f(-1) = -1 + 4 + 7 - 10
We'll combine like terms and we'll have:
f(-1) = 0
If f(-1) = 0, then x = -1 is the root of polynomial.
We'll write the rule of division with reminder:
f(x) = (x + 1)(ax^2 + bx + c) + 0
The reminder is 0, since f(-1) = 0
x^3+4x^2-7x-10 = (x + 1)(ax^2 + bx + c)
We'll remove the brackets:
x^3+4x^2-7x-10 = ax^3 + bx^2 + cx + ax^2 + bx + c
Comparing, we'll get:
a = 1
a + b = 4
b = 4 - a
b = 3
b + c = -7
c = -7 - b
c = -10
ax^2 + bx + c = x^2 + 3x - 10
Now, we'l determine the roots of the quotient x^2 + 3x - 10.
x^2 + 3x - 10 = 0
x1 = [-3+sqrt(9 + 40)]/2
x1 = (-3+7)/2
x1 = 2
x2 = -5
The complete factorized polynomial is:
f(x) = (x + 1)(x - 2)(x + 5)
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