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MathematicsPrecalculusQuadratic Equations → Factor Quadratic Equations

Instructions for problem set  Factor each quadratic equation

Questions


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

\(y = x^2 -2 x -24\)

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

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

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

\(y = x^2 -3 x -10\)

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

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

\(y = 3 x^2 + 8 x -3\)

\(y = x^2 + 4 x -12\)

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

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

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

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

\(y = 4 x^2 -7 x -2\)

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

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

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

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

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

\(y = 3 x^2 -5 x -2\)

\(y = x^2 -x -20\)

\(y = x^2 + 2 x + 1\)

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

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

\(y = x^2 + 2 x -24\)

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

\(y = 3 x^2 -10 x + 3\)

Answers


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

\(y = x^2 -2 x -24 \) ⇒
\((x -6)(x + 4)\)

\(y = x^2 -5 x \) ⇒
\((x -5)(x)\)

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

\(y = 2 x^2 -8 x + 6 \) ⇒
\(2(x -1)(x -3)\)

\(y = x^2 -3 x -10 \) ⇒
\((x + 2)(x -5)\)

\(y = x^2 -6 x + 5 \) ⇒
\((x -5)(x -1)\)

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

\(y = 3 x^2 + 8 x -3 \) ⇒
\((3 x -1)(x + 3)\)

\(y = x^2 + 4 x -12 \) ⇒
\((x + 6)(x -2)\)

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

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

\(y = x^2 -6 x \) ⇒
\((x)(x -6)\)

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

\(y = 4 x^2 -7 x -2 \) ⇒
\((4 x + 1)(x -2)\)

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

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

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

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

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

\(y = 3 x^2 -5 x -2 \) ⇒
\((3 x + 1)(x -2)\)

\(y = x^2 -x -20 \) ⇒
\((x + 4)(x -5)\)

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

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

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

\(y = x^2 + 2 x -24 \) ⇒
\((x -4)(x + 6)\)

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

\(y = 3 x^2 -10 x + 3 \) ⇒
\((3 x -1)(x -3)\)
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