Question

Factorize using factor theorem 6x^3-x^2-12x-5​

Answer 1

Answer:

(x+1)(2x+1)(3x-5)

Step-by-step explanation:

Let p(x) =6x³-x²-12x-5

Now by trial and error method, we get

p(-1)=6(-1)³-(-1)²-12(-1)-5=-6-1+12-5=0

∴ by factor theorem (x+1)|p(x) [ | means "factor of"]

So, p(x) is divisible by (x+1)

(6x³-x²-12x-5)/(x-1)=6x²-7x-5

=>(6x³-x²-12x-5)/(x-1)=(2x+1)(3x-5)

=>6x³-x²-12x-5=(x-1)(2x+1)(3x-5)

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Answer 2

Step-by-step explanation:

first check that by taking what the result becomes zero of p(x)...

so,taking x=-1 the p(x) cones zero...

it means (x+1) is a factor of p(x)

then divide the p(x) by (x+1)

the quotient comes 6x²-7x-5

then take (x+1)(6x²-7x-5)

=(x+1){6x²-(10-3)x-5}

=(x+1){6x²-10x+3x-5}

=(x+1){2x(3x-5)+1(3x-5)}

=(x+1)(2x+1)(3x-5)