Question

Factorise 2x^3 - 5x^2 - 19x + 4​

Answer 1

Step-by-step explanation:

Answer:

$\huge\underline\textsf{Question:- }$

$\boxed{\sf2x ^{3} - 5x {}^{2} - 19x + 42}$

$\huge\underline\textsf{Explantion:- }$

$\leadsto\sf\red{2x {}^{3} - 4x {}^{2} - x {}^{2} + 2x - 21x + 42}$

$\leadsto\sf\orange{2x {}^{2} (x - 2) - x(x - 2) - 21(x - 2)}$

$\leadsto\sf\purple{(x - 2)(2x {}^{2} - x - 2)}$

$\leadsto\sf\blue{(x - 2)(2x {}^{2} + 6x - 7x - 21)}$

$\leadsto\sf\pink{(x - 2)(2x {}^{2} + 6x - 7x - 21)}$

$\leadsto\sf\orange{(x - 2)[(2x(x + 3) - 7(x + 3)]}$

$\leadsto\sf\red{(x - 2)(2x - 7)(x + 3)}$

$\large\underline\textsf{\green{Ans.(x - 2)(2x - 7)(x + 3)} }$

Answer 2

We can use synthetic division to look for factors:

The possible zeroes are the (factors of 42)/(factors of 2)

Possible zeroes: ±{1, 1/2, 3, 3/2, 6, 7, 7/2, 14, 21, 10 1/2, 42}

If (x+1) is a factor then x = -1 is a zero:

-1 | 2 -5 -19 42

| -2 7 12

----------------------

2 -7 -12 54 This has remainder ≠0 so not a factor

I also did (x-1), (x-2), (x+2) until I got to (x+3)

-3 | 2 -5 -19 42

| -6 33 -42

--------------------------

2 -11 14 0 This has remainder of zero so -3 is a zero

meaning that x+3 is a factor

2x^3-5x^2-19x+42 = (x+3)(2x2 - 11x + 14)

We can factor the 2x2 - 11x + 14 by using grouping:

Looking at the coefficients of the x2 term and the constant

we have 2 & 14... 2(14) = 28. Look for factors of 28 that

add to the coefficient of the x term ... -11

(-4)(-7) = 28 and -4-7 = -11.

Replace -11x with -4x - 7x and factor by grouping

2x2 - 11x + 14 = 2x2 - 4x - 7x + 14

= 2x(x - 2) -7(x - 2) = (2x-7)(x-2)

Our final answer:

2x3-5x2-19x+42 = (x+3)(2x-7)(x-2)