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Mathematics → Precalculus → Quadratic Equations → Factor Quadratic Equations

Instructions for problem set  Factor each quadratic equation

Questions


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

\(y = 2 x^2 + 7 x + 6\)

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

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

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

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

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

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

\(y = x^2 + 7 x + 12\)

\(y = x^2 + 11 x + 30\)

\(y = x^2 -7 x + 12\)

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

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

\(y = x^2 -10 x + 24\)

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

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

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

\(y = x^2 + 8 x + 15\)

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

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

\(y = x^2 + 9 x + 18\)

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

\(y = 4 x^2 + 9 x -9\)

\(y = x^2 -7 x + 10\)

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

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

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

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

Answers


\(y = x^2 + 8 x + 12 \) ⇒
\((x + 2)(x + 6)\)
\(y = 2 x^2 + 7 x + 6 \) ⇒
\((2 x + 3)(x + 2)\)
\(y = 3 x^2 -x \) ⇒
\((3 x -1)(x)\)
\(y = x^2 -4 x + 3 \) ⇒
\((x -3)(x -1)\)

\(y = x^2 -3 x -18 \) ⇒
\((x -6)(x + 3)\)
\(y = x^2 + x -6 \) ⇒
\((x + 3)(x -2)\)
\(y = x^2 + 3 x -18 \) ⇒
\((x + 6)(x -3)\)
\(y = 4 x^2 + x \) ⇒
\((4 x + 1)(x)\)

\(y = x^2 + 7 x + 12 \) ⇒
\((x + 3)(x + 4)\)
\(y = x^2 + 11 x + 30 \) ⇒
\((x + 5)(x + 6)\)
\(y = x^2 -7 x + 12 \) ⇒
\((x -3)(x -4)\)
\(y = x^2 -x -30 \) ⇒
\((x + 5)(x -6)\)

\(y = 2 x^2 -2 \) ⇒
\(2(x + 1)(x -1)\)
\(y = x^2 -10 x + 24 \) ⇒
\((x -4)(x -6)\)
\(y = x^2 -4 x -5 \) ⇒
\((x -5)(x + 1)\)
\(y = 3 x^2 + 3 x \) ⇒
\(3(x)(x + 1)\)

\(y = x^2 -2 x -8 \) ⇒
\((x -4)(x + 2)\)
\(y = x^2 + 8 x + 15 \) ⇒
\((x + 5)(x + 3)\)
\(y = x^2 + 3 x -10 \) ⇒
\((x + 5)(x -2)\)
\(y = x^2 -5 x \) ⇒
\((x -5)(x)\)

\(y = x^2 + 9 x + 18 \) ⇒
\((x + 6)(x + 3)\)
\(y = x^2 -x -12 \) ⇒
\((x -4)(x + 3)\)
\(y = 4 x^2 + 9 x -9 \) ⇒
\((4 x -3)(x + 3)\)
\(y = x^2 -7 x + 10 \) ⇒
\((x -5)(x -2)\)

\(y = 4 x^2 + 4 x \) ⇒
\(4(x)(x + 1)\)
\(y = x^2 + 3 x -4 \) ⇒
\((x -1)(x + 4)\)
\(y = 2 x^2 -6 x \) ⇒
\(2(x)(x -3)\)
\(y = x^2 -5 x + 6 \) ⇒
\((x -2)(x -3)\)
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