factorise : x^3-5x^2-2x+4
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Use the rational root theorem to get started, then factor the remaining quadratic to find:
x3−5x2−2x+24=(x+2)(x−4)(x−3)
Explanation:
Let f(x)=x3−5x2−2x+24
By the rational root theorem, any rational zeros of f(x) must be expressible in the for pq for integers p, q with p a divisor of the constant term 24 and q a divisor of the coefficient 1 of the leading term.
That means that the only possible rational zeros are the factors of 24, namely:
±1,±2,±3,±4,±6,±12,±24
Try each in turn:
f(1)=1−5−2+24=18
f(−1)=−1−5+2+24=20
f(2)=8−20−4+24=8
f(−2)=−8−20+4+24=0
So x=−2 is a zero and (x+2) is a factor.
x3−5x2−2x+24=(x+2)(x2−7x+12)
We can factor x2−7x+12 by noting that 4×3=12 and 4+3=7, so:
x2−7x+12=(x−4)(x−3)
Putting it all together:
x3−5x2−2x+24=(x+2)(x−4)(x−3)
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x3−5x2−2x+24=(x+2)(x−4)(x−3)
Explanation:
Let f(x)=x3−5x2−2x+24
By the rational root theorem, any rational zeros of f(x) must be expressible in the for pq for integers p, q with p a divisor of the constant term 24 and q a divisor of the coefficient 1 of the leading term.
That means that the only possible rational zeros are the factors of 24, namely:
±1,±2,±3,±4,±6,±12,±24
Try each in turn:
f(1)=1−5−2+24=18
f(−1)=−1−5+2+24=20
f(2)=8−20−4+24=8
f(−2)=−8−20+4+24=0
So x=−2 is a zero and (x+2) is a factor.
x3−5x2−2x+24=(x+2)(x2−7x+12)
We can factor x2−7x+12 by noting that 4×3=12 and 4+3=7, so:
x2−7x+12=(x−4)(x−3)
Putting it all together:
x3−5x2−2x+24=(x+2)(x−4)(x−3)
pls mark as brainliest pls pls pls
rmn24:
thank u a lot
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(X+1)(x^2+6X+4)
it's right answer.
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