Is it possible to factor $y = \frac{1}{2} x^{2} + \frac{3}{2} x - 4$ $y= 1/2x^2 + 3/2x -4$ ? If so, what are the factors?

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Answer 1 · 2018-06-12T18:26:14

Yes
$(x + 4) \times (x - 1)$ $( x +4) xx ( x -1)$

first consider y to be any number and its value not needed for the purpose of factoring

second multiply everything by 2

$2 {(\frac{1}{2} x^{2} + (\frac{3}{2}) x - 4}$ $2 { (1/2 x^2 + (3/2) x - 4}$ = $x^{2} + 3 x - 4$ $x^2 + 3x -4$

Now the x terms are in the standard $A x^{2} + B x + C$ $Ax^2 + Bx + C$

Since the C is -4 one binomial factor must be negative and the other binomial factor must be positive.

The B is +3 which means the positive binomial factor must be greater than the the negative binomial factor.

third used the information given by the B and C values to find the binomial factors.

The only factors of 4 are $2 \times 2$ $2 xx 2$ and $4 \times 1$ $4 xx 1$

$4 - 1 = 3$ $4- 1 = 3$

The binomial factors then are

$(x + 4) \times (x - 1)$ $( x +4) xx ( x -1)$

Is it possible to factor y=12x2+32x−4y= 1/2x^2 + 3/2x -4 ? If so, what are the factors?