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MathematicsPrecalculusQuadratic Equations → Factor Quadratic Equations

Instructions for problem set  Factor each quadratic equation

Questions


\(y = 3 x^2\)

\(y = 3 x^2 + 9 x + 6\)

\(y = 4 x^2 + 6 x + 2\)

\(y = 4 x^2 -14 x + 6\)

\(y = 3 x^2 + 2 x\)

\(y = x^2 -5 x + 6\)

\(y = 3 x^2 -2 x\)

\(y = 2 x^2 + 2 x -4\)

\(y = 2 x^2 -4 x\)

\(y = x^2 -10 x + 25\)

\(y = x^2 + 2 x -3\)

\(y = x^2 + x -6\)

\(y = x^2 + 3 x\)

\(y = x^2 -x -30\)

\(y = x^2 + 4 x + 3\)

\(y = 2 x^2 -3 x -2\)

\(y = x^2 + 6 x + 8\)

\(y = 4 x^2 + 2 x -2\)

\(y = 4 x^2 -5 x + 1\)

\(y = x^2 + 8 x + 12\)

\(y = 2 x^2\)

\(y = 4 x^2 -13 x + 3\)

\(y = x^2 -3 x -18\)

\(y = 2 x^2 -7 x + 3\)

\(y = x^2 + 5 x\)

\(y = 3 x^2 -3\)

\(y = x^2 + 5 x + 6\)

\(y = 4 x^2 -10 x + 4\)

Answers


\(y = 3 x^2 \) ⇒
\(3(x)(x)\)

\(y = 3 x^2 + 9 x + 6 \) ⇒
\(3(x + 1)(x + 2)\)

\(y = 4 x^2 + 6 x + 2 \) ⇒
\(2(2 x + 1)(x + 1)\)

\(y = 4 x^2 -14 x + 6 \) ⇒
\(2(2 x -1)(x -3)\)

\(y = 3 x^2 + 2 x \) ⇒
\((3 x + 2)(x)\)

\(y = x^2 -5 x + 6 \) ⇒
\((x -2)(x -3)\)

\(y = 3 x^2 -2 x \) ⇒
\((3 x -2)(x)\)

\(y = 2 x^2 + 2 x -4 \) ⇒
\(2(x -1)(x + 2)\)

\(y = 2 x^2 -4 x \) ⇒
\(2(x)(x -2)\)

\(y = x^2 -10 x + 25 \) ⇒
\((x -5)(x -5)\)

\(y = x^2 + 2 x -3 \) ⇒
\((x + 3)(x -1)\)

\(y = x^2 + x -6 \) ⇒
\((x + 3)(x -2)\)

\(y = x^2 + 3 x \) ⇒
\((x + 3)(x)\)

\(y = x^2 -x -30 \) ⇒
\((x -6)(x + 5)\)

\(y = x^2 + 4 x + 3 \) ⇒
\((x + 3)(x + 1)\)

\(y = 2 x^2 -3 x -2 \) ⇒
\((2 x + 1)(x -2)\)

\(y = x^2 + 6 x + 8 \) ⇒
\((x + 4)(x + 2)\)

\(y = 4 x^2 + 2 x -2 \) ⇒
\(2(2 x -1)(x + 1)\)

\(y = 4 x^2 -5 x + 1 \) ⇒
\((4 x -1)(x -1)\)

\(y = x^2 + 8 x + 12 \) ⇒
\((x + 6)(x + 2)\)

\(y = 2 x^2 \) ⇒
\(2(x)(x)\)

\(y = 4 x^2 -13 x + 3 \) ⇒
\((4 x -1)(x -3)\)

\(y = x^2 -3 x -18 \) ⇒
\((x -6)(x + 3)\)

\(y = 2 x^2 -7 x + 3 \) ⇒
\((2 x -1)(x -3)\)

\(y = x^2 + 5 x \) ⇒
\((x + 5)(x)\)

\(y = 3 x^2 -3 \) ⇒
\(3(x -1)(x + 1)\)

\(y = x^2 + 5 x + 6 \) ⇒
\((x + 2)(x + 3)\)

\(y = 4 x^2 -10 x + 4 \) ⇒
\(2(2 x -1)(x -2)\)
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