How do you factor the expression $2 x^{2} + 7 x + 1$ $2x^2+7x+1$ ?

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How do you factor the expression $2 x^{2} + 7 x + 1$ $2x^2+7x+1$ ?

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Answer 1 · 2018-03-30T15:36:59

the answer is $\frac{- 7 \sqrt{41}}{4}$ $(-7sqrt(41))/4$

Explanation:

you can't take out so you have to use the quadratic formula. formula: $\frac{- b \sqrt{b^{2} - 4 a c}}{2 a}$ $(-bsqrt(b^2-4ac))/(2a)$ so plug it in wiht the numbers: $- 7 \frac{\sqrt{{(7)}^{2} - 4 (2) (1)}}{2 (2)}$ $-7sqrt((7)^2-4(2)(1))/(2(2))$ now solve so you'll get: $- 7 \frac{\sqrt{49 - 8}}{4}$ $-7sqrt(49-8)/(4)$ now you can simplify $41$ $41$ anymore so you leave it inside. and there you go. $41$ $41$ comes from $49 - 8$ $49-8$

Answer 2 · 2018-03-30T15:37:48

check below

Explanation:

$2 x^{2}$ $2x^2$ + $7 x$ $7x$ + 1
Since you can not use middle-term techniques over here.. So u can use the formula of discriminant:

myself

The equation is :

$a x^{2}$ $ax^2$ + $b x$ $bx$ +c
Here a,b,c are the coefficients of $x^{2}$ $x^2$ , $x$ $x$ and the constant respectively,,,

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How do you factor the expression $2 x^{2} + 7 x + 1$ $2x^2+7x+1$ ?

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