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Question

# Factorise. (1) x2 + 9x + 18 (2) x2 − 10x + 9 (3) y2 + 24y + 144 (4) 5y2 + 5y − 10 (5) p2 − 2p − 35 (6) p2 − 7p − 44 (7) m2 − 23m + 120 (8) m2 − 25m + 100 (9) 3x2 + 14x + 15 (10) 2x2 + x − 45 (11) 20x2 − 26x + 8 (12) 44x2 − x − 3

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Solution

## $\left(1\right){x}^{2}+9x+18\phantom{\rule{0ex}{0ex}}={x}^{2}+6x+3x+18\phantom{\rule{0ex}{0ex}}=x\left(x+6\right)+3\left(x+6\right)\phantom{\rule{0ex}{0ex}}=\left(x+3\right)\left(x+6\right)$ $\left(2\right){x}^{2}-10x+9\phantom{\rule{0ex}{0ex}}={x}^{2}-9x-x+9\phantom{\rule{0ex}{0ex}}=x\left(x-9\right)-1\left(x-9\right)\phantom{\rule{0ex}{0ex}}=\left(x-1\right)\left(x-9\right)$ $\left(3\right){y}^{2}+24y+144\phantom{\rule{0ex}{0ex}}={y}^{2}+12y+12y+144\phantom{\rule{0ex}{0ex}}=y\left(y+12\right)+12\left(y+12\right)\phantom{\rule{0ex}{0ex}}=\left(y+12\right)\left(y+12\right)$ $\left(4\right)5{y}^{2}+5y-10\phantom{\rule{0ex}{0ex}}=5{y}^{2}+10y-5y-10\phantom{\rule{0ex}{0ex}}=5y\left(y+2\right)-5\left(y+2\right)\phantom{\rule{0ex}{0ex}}=\left(5y-5\right)\left(y+2\right)\phantom{\rule{0ex}{0ex}}=5\left(y-1\right)\left(y+2\right)$ $\left(5\right){p}^{2}-2p-35\phantom{\rule{0ex}{0ex}}={p}^{2}-7p+5p-35\phantom{\rule{0ex}{0ex}}=p\left(p-7\right)+5\left(p-7\right)\phantom{\rule{0ex}{0ex}}=\left(p+5\right)\left(p-7\right)$ $\left(6\right){p}^{2}-7p-44\phantom{\rule{0ex}{0ex}}={p}^{2}-11p+4p-44\phantom{\rule{0ex}{0ex}}=p\left(p-11\right)+4\left(p-11\right)\phantom{\rule{0ex}{0ex}}=\left(p+4\right)\left(p-11\right)$ $\left(7\right){m}^{2}-23m+120\phantom{\rule{0ex}{0ex}}={m}^{2}-15m-8m+120\phantom{\rule{0ex}{0ex}}=m\left(m-15\right)-8\left(m-15\right)\phantom{\rule{0ex}{0ex}}=\left(m-8\right)\left(m-15\right)$ $\left(8\right){m}^{2}-25m+100\phantom{\rule{0ex}{0ex}}={m}^{2}-20m-5m+100\phantom{\rule{0ex}{0ex}}=m\left(m-20\right)-5\left(m-20\right)\phantom{\rule{0ex}{0ex}}=\left(m-5\right)\left(m-20\right)$ $\left(9\right)3{x}^{2}+14x+15\phantom{\rule{0ex}{0ex}}=3{x}^{2}+9x+5x+15\phantom{\rule{0ex}{0ex}}=3x\left(x+3\right)+5\left(x+3\right)\phantom{\rule{0ex}{0ex}}=\left(3x+5\right)\left(x+3\right)$ $\left(10\right)2{x}^{2}+x-45\phantom{\rule{0ex}{0ex}}=2{x}^{2}+10x-9x-45\phantom{\rule{0ex}{0ex}}=2x\left(x+5\right)-9\left(x+5\right)\phantom{\rule{0ex}{0ex}}=\left(2x-9\right)\left(x+5\right)$ $\left(11\right)20{x}^{2}-26x+8\phantom{\rule{0ex}{0ex}}=20{x}^{2}-16x-10x+8\phantom{\rule{0ex}{0ex}}=4x\left(5x-4\right)-2\left(5x-4\right)\phantom{\rule{0ex}{0ex}}=\left(4x-2\right)\left(5x-4\right)\phantom{\rule{0ex}{0ex}}=2\left(2x-1\right)\left(5x-4\right)$ $\left(12\right)44{x}^{2}-x-3\phantom{\rule{0ex}{0ex}}=44{x}^{2}+11x-12x-3\phantom{\rule{0ex}{0ex}}=11x\left(4x+1\right)-3\left(4x+1\right)\phantom{\rule{0ex}{0ex}}=\left(11x-3\right)\left(4x+1\right)$

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