Quadratic polynomial deserve to be factored using the change ax^2+bx+c=a\left(x-x_1\right)\left(x-x_2\right), whereby x_1 and x_2 room the services of the quadratic equation ax^2+bx+c=0.

You are watching: X^2-14x+49

every equations that the type ax^2+bx+c=0 have the right to be resolved using the quadratic formula: \frac-b±\sqrtb^2-4ac2a. The quadratic formula offers two solutions, one once ± is addition and one as soon as it is subtraction.
aspect the original expression using ax^2+bx+c=a\left(x-x_1\right)\left(x-x_2\right). Substitute -7+7\sqrt2 because that x_1 and also -7-7\sqrt2 because that x_2.

https://socratic.org/questions/how-do-you-use-the-important-points-to-sketch-the-graph-of-x-2-14x-49
graph-x^2+14x-49 <-10, 10, -5, 5> Explanation:The general type for a quadratic equation is\displaystyley=ax^2+bx+cIn this question, the equation you wish to graph must be ...
\displaystyle\left(x+7\right)^2=0 Explanation: \displaystylex^2+14x=-49\mid+49\displaystylex^2+14x+49=0\displaystyle\textComplete the perfect square ...
5x2+14x-9 Final an outcome : 5x2 + 14x - 9 action by step solution : action 1 :Equation at the end of action 1 : (5x2 + 14x) - 9 step 2 :Trying to element by dividing the middle term 2.1 Factoring ...
3x2+14x-49=0 Two services were found : x = -7 x = 7/3 = 2.333 step by action solution : step 1 :Equation in ~ the end of step 1 : (3x2 + 14x) - 49 = 0 action 2 :Trying to factor by dividing ...
\displaystyley=\left(5x-1\right)\left(x+3\right) Explanation:We have\displaystyley=5x^2+14x-3 We begin by multiplying\displaystyle5and\displaystyle-3 ...
More Items

Quadratic polynomial have the right to be factored using the change ax^2+bx+c=a\left(x-x_1\right)\left(x-x_2\right), where x_1 and x_2 space the remedies of the quadratic equation ax^2+bx+c=0.
All equations that the type ax^2+bx+c=0 have the right to be resolved using the quadratic formula: \frac-b±\sqrtb^2-4ac2a. The quadratic formula offers two solutions, one as soon as ± is enhancement and one as soon as it is subtraction.

See more: How Are Vascular Bundles Arranged In A Monocot Stem And Dicot Stem: Difference

Factor the initial expression utilizing ax^2+bx+c=a\left(x-x_1\right)\left(x-x_2\right). Substitute -7+7\sqrt2 because that x_1 and also -7-7\sqrt2 because that x_2.
\left< \beginarray l l 2 & 3 \\ 5 & 4 \endarray \right> \left< \beginarray together l together 2 & 0 & 3 \\ -1 & 1 & 5 \endarray \right>

EnglishDeutschEspañolFrançaisItalianoPortuguêsРусский简体中文繁體中文Bahasa MelayuBahasa Indonesiaالعربية日本語TürkçePolskiעבריתČeštinaNederlandsMagyar Nyelv한국어SlovenčinaไทยελληνικάRomânăTiếng Việtहिन्दीঅসমীয়াবাংলাગુજરાતીಕನ್ನಡकोंकणीമലയാളംमराठीଓଡ଼ିଆਪੰਜਾਬੀதமிழ்తెలుగు