Resolve each of the following quadratic trinomial into factor:
28 − 31x − 5x2
The given expression is $28-31 \mathrm{x}-5 \mathrm{x}^{2}$. (Coefficient of $\mathrm{x}^{2}=-5$, coefficient of $\mathrm{x}=-31$ and constant term $=28)$
We will split the coefficient of $\mathrm{x}$ into two parts such that their sum is $-31$ and their product equals the product of the coefficient of $\mathrm{x}^{2}$ and the constant term, i.e., $(-5) \times(28)=-140$.
Now,
$(-35)+4=-31$
and
$(-35) \times 4=-140$
Replacing the middle term $-31 \mathrm{x}$ by $-35 \mathrm{x}+4 \mathrm{x}$, we have :
$-5 x^{2}-31 x+28=-5 x^{2}-35 x+4 x+28$
$=\left(-5 x^{2}-35 x\right)+(4 x+28)$
$=-5 x(x+7)+4(x+7)$
$=(4-5 x)(x+7)$
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